B.1 Complex numbers · Matrix Theory

Section 1.1.4 built $ℂ$ as $ℝ^{2}$ with a new multiplication and proved that it is a field. This appendix collects what else the book uses about complex numbers. Conjugation and the modulus are the raw material of inner products over $ℂ$ (Chapter 6). Polar form explains what multiplying by a complex number does to the plane. Roots of unity give the discrete Fourier basis (Section 6.3.4), and square roots give the first cases of the Fundamental Theorem of Algebra (Theorem 7.1.32).

Throughout, when we write $𝑧 = 𝑎 + 𝑏 𝑖$ or $𝑤 = 𝑐 + 𝑑 𝑖$ , the letters $𝑎, 𝑏, 𝑐, 𝑑$ denote real numbers.

B.1.1Conjugation

The multiplication of $ℂ$ is forced by the single rule $𝑖^{2} = - 1$ (Section 1.1.4). But ${(- 𝑖)}^{2} = - 1$ as well, and nothing in that rule, or in the arithmetic of $ℝ$ , prefers $𝑖$ to $- 𝑖$ . Exchanging the two is conjugation.

Definition B.1.1 (Complex conjugate). Let

𝑧 = 𝑎 + 𝑏 𝑖 \in ℂ

. The complex conjugate of

𝑧

𝑧 = 𝑎 - 𝑏 𝑖

. We write

Re 𝑧 = 𝑎

and

Im 𝑧 = 𝑏

for the real and imaginary parts of

𝑧

(Definition 1.1.29); both are real numbers, and

𝑧 = Re 𝑧 + 𝑖 Im 𝑧

. We call

𝑧

purely imaginary if

Re 𝑧 = 0

For example, $3 + 2 𝑖 = 3 - 2 𝑖$ , $𝑖 = - 𝑖$ , and $5 = 5$ . In the plane, $𝑧$ is the mirror image of $𝑧$ in the real axis (Figure B.1.1). As a map $ℂ \to ℂ$ , conjugation is $ℝ$ -linear but not $ℂ$ -linear (Non-example 2.1.14).

For $𝑧 \neq 0$ and an integer $𝑛 > 0$ we write $𝑧^{- 𝑛} = {(𝑧^{- 1})}^{𝑛}$ . The laws $𝑧^{𝑚} 𝑧^{𝑛} = 𝑧^{𝑚 + 𝑛}$ , ${(𝑧^{𝑚})}^{𝑛} = 𝑧^{𝑚 𝑛}$ , and ${(𝑧 𝑤)}^{𝑛} = 𝑧^{𝑛} 𝑤^{𝑛}$ then hold for all integers $𝑚, 𝑛$ (with $𝑧, 𝑤 \neq 0$ when an exponent is negative), by the same inductions as for real numbers, and we use them freely.

Proposition B.1.2 (Conjugation respects the field operations). Let $𝑧, 𝑤 \in ℂ$ .

(a) $𝑧 + 𝑤 = 𝑧 + 𝑤$ and $𝑧 𝑤 = 𝑧 𝑤$ .
(b) $𝑧 = 𝑧$ . In particular, conjugation is a bijection $ℂ \to ℂ$ and is its own inverse.
(c) $𝑧 = 𝑧$ if and only if $𝑧 \in ℝ$ , and $𝑧 = - 𝑧$ if and only if $𝑧$ is purely imaginary.
(d) $Re 𝑧 = (𝑧 + 𝑧) / 2$ and $Im 𝑧 =$ $(𝑧 - 𝑧) / (2 𝑖)$ .
(e) $- 𝑧 = - 𝑧$ . If $𝑧 \neq 0$ , then $𝑧 \neq 0$ and $𝑧^{- 1} = 𝑧^{- 1}$ . Consequently $𝑧^{𝑛} = 𝑧^{𝑛}$ for every integer $𝑛$ (with $𝑧 \neq 0$ if $𝑛 < 0$ ).

Proof.routine

Uses. Definition B.1.1 (Complex conjugate), Definition 1.1.29 (Complex numbers), Theorem 1.1.9 (Uniqueness of identities and inverses), induction on $𝑛$

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Write

𝑧 = 𝑎 + 𝑏 𝑖

and

𝑤 = 𝑐 + 𝑑 𝑖

, and compute both sides of (a)–(d) from Definition 1.1.29.

Hint 2. For (e), avoid coordinates. Apply (a) to

𝑧 + (- 𝑧) = 0

and to

𝑧 𝑧^{- 1} = 1

, and use the uniqueness of inverses. Then induct on

𝑛

Sketch. Parts (a)–(d) are coordinate computations; for the product, both sides equal

(𝑎 𝑐 - 𝑏 𝑑) -

(𝑎 𝑑 + 𝑏 𝑐) 𝑖

. For (e), conjugating

𝑧 + (- 𝑧) = 0

and

𝑧 𝑧^{- 1} = 1

with (a) shows that

- 𝑧

and

𝑧^{- 1}

are the additive and multiplicative inverses of

𝑧

. Powers follow from (a) by induction.

Let $𝑧 = 𝑎 + 𝑏 𝑖$ and $𝑤 = 𝑐 + 𝑑 𝑖$ .

(a) $𝑧 + 𝑤 =$ $(𝑎 + 𝑐) -$ $(𝑏 + 𝑑) 𝑖 =$ $(𝑎 - 𝑏 𝑖) +$ $(𝑐 - 𝑑 𝑖) =$ $𝑧 +$ $𝑤$ . Since $𝑧 𝑤 =$ $(𝑎 𝑐 - 𝑏 𝑑) +$ $(𝑎 𝑑 + 𝑏 𝑐) 𝑖$ ,

𝑧 𝑤 = (𝑎 𝑐 - 𝑏 𝑑) - (𝑎 𝑑 + 𝑏 𝑐) 𝑖 = (𝑎 - 𝑏 𝑖) (𝑐 - 𝑑 𝑖) = 𝑧 𝑤,

where the middle equality is the multiplication rule of Definition 1.1.29 applied to $𝑎 + (- 𝑏) 𝑖$ and $𝑐 + (- 𝑑) 𝑖$ : the real part of the product is $𝑎 𝑐 -$ $(- 𝑏) (- 𝑑) =$ $𝑎 𝑐 -$ $𝑏 𝑑$ , and its imaginary part is $𝑎 (- 𝑑) +$ $(- 𝑏) 𝑐 =$ $- (𝑎 𝑑 + 𝑏 𝑐)$ .

(b) $𝑧 =$ $𝑎 - 𝑏 𝑖 =$ $𝑎 +$ $𝑏 𝑖 =$ $𝑧$ . A map that is its own inverse is a bijection.

(c) $𝑧 = 𝑧$ means $- 𝑏 = 𝑏$ , that is, $𝑏 = 0$ , which says that $𝑧 = 𝑎 \in ℝ$ . Likewise $𝑧 = - 𝑧$ means $𝑎 = - 𝑎$ , that is, $Re 𝑧 = 0$ .

(d) $𝑧 + 𝑧 = 2 𝑎$ and $𝑧 - 𝑧 = 2 𝑏 𝑖$ .

(e) Since $𝑧 + (- 𝑧) = 0$ and $0 = 0$ , part (a) gives $𝑧 + - 𝑧 = 0$ , so $- 𝑧 = - 𝑧$ by Theorem 1.1.9. If $𝑧 \neq 0$ , then $𝑧 𝑧^{- 1} = 1$ and $1 = 1$ , so (a) gives $𝑧 𝑧^{- 1} = 1$ . Hence $𝑧 \neq 0$ , and $𝑧^{- 1} = 𝑧^{- 1}$ by Theorem 1.1.9.

For $𝑛 \geq 0$ we show $𝑧^{𝑛} = 𝑧^{𝑛}$ by induction on $𝑛$ . The case $𝑛 = 0$ is $1 = 1$ , and if it holds for $𝑛$ , then by (a)

𝑧^{𝑛 + 1} = 𝑧^{𝑛} 𝑧 = 𝑧^{𝑛} 𝑧 = 𝑧^{𝑛} 𝑧 = 𝑧^{𝑛 + 1} .

For $𝑛 < 0$ and $𝑧 \neq 0$ , the case $- 𝑛 > 0$ applied to $𝑧^{- 1}$ gives $𝑧^{𝑛} =$ ${(𝑧^{- 1})}^{- 𝑛} =$ ${𝑧^{- 1}}^{- 𝑛} =$ ${(𝑧^{- 1})}^{- 𝑛} =$ $𝑧^{𝑛}$ . □

A bijection $𝜎 : 𝐹 \to 𝐹$ of a field with $𝜎 (𝑎 + 𝑏) =$ $𝜎 (𝑎) +$ $𝜎 (𝑏)$ and $𝜎 (𝑎 𝑏) =$ $𝜎 (𝑎) 𝜎 (𝑏)$ for all $𝑎, 𝑏 \in 𝐹$ is an automorphism of $𝐹$ . By (a) and (b), conjugation is an automorphism of $ℂ$ , and by (c) the numbers it fixes are exactly the real ones. Exercise B.1.43 shows that it is the only automorphism of $ℂ$ other than the identity that fixes every real number.

Intuition (Conjugation as a symmetry). Read Proposition B.1.2 as a symmetry of

ℂ

. Take any true equation built from complex numbers with sums, products, and inverses. Conjugate both sides and push the bars inward with (a) and (e): every complex number in it is replaced by its conjugate, and every real number stays as it is. So the new equation is also true. This is the precise sense in which

𝑖

and

- 𝑖

cannot be told apart.

Technique B.1.3 (Conjugate the equation). Conjugation respects sums, products, and inverses, and it fixes real numbers (Proposition B.1.2). So conjugating both sides of a true equation gives another true equation, in which each complex number is replaced by its conjugate and each real number is unchanged. Two uses come up again and again.

To get a second solution of an equation with real coefficients from a first one, conjugate the first. The canonical example is Proposition B.1.4: the nonreal roots of a real polynomial come in pairs $𝑧, 𝑧$ .
To show that a complex quantity $𝑞$ is real, show that $𝑞 = 𝑞$ ; to show that it is purely imaginary, show that $𝑞 = - 𝑞$ (Proposition B.1.2(c)).

Reach for it when real coefficients meet complex unknowns, as when a real matrix acts on $ℂ^{𝑛}$ , or when a quantity built symmetrically from $𝑧$ and $𝑧$ ought to be real. The second use shows that the eigenvalues of a self-adjoint operator are real (Proposition 8.1.1).

Used in 3 places and 3 practice exercises

Problem 7.18, Proposition B.1.4, Exercise B.1.40; 3 practice exercises.

Proposition B.1.4 (Roots of real polynomials come in conjugate pairs). Let

𝑝 =

𝑎_{0} +

𝑎_{1} 𝑥 +

\dots +

𝑎_{𝑑} 𝑥^{𝑑}

with

𝑎_{0}, \dots, 𝑎_{𝑑} \in ℝ

, and evaluate

𝑝

at complex numbers (Definition 1.2.17, with

𝐹 = ℂ

). Then

𝑝 (𝑧) = 𝑝 (𝑧)

for every

𝑧 \in ℂ

. In particular, if

𝑝 (𝑧) = 0

, then

𝑝 (𝑧) = 0

Proof.routine

Uses. Proposition B.1.2 (Conjugation respects the field operations), Definition 1.2.17 (Evaluation)

Techniques. Conjugate the equation (Technique B.1.3)

Hint 1. Conjugate the number

𝑝 (𝑧) =

\sum_{𝑘} 𝑎_{𝑘} 𝑧^{𝑘}

and push the bar inward.

Hint 2. The bar passes through sums and products by Proposition B.1.2(a), and

𝑎_{𝑘} = 𝑎_{𝑘}

because

𝑎_{𝑘}

is real.

Sketch.

𝑝 (𝑧) =

\sum_{𝑘} 𝑎_{𝑘} 𝑧^{𝑘} =

\sum_{𝑘} 𝑎_{𝑘} 𝑧^{𝑘} =

𝑝 (𝑧)

. If

𝑝 (𝑧) = 0

, then

𝑝 (𝑧) = 0 = 0

By Proposition B.1.2(a) and (e), and since $𝑎_{𝑘} = 𝑎_{𝑘}$ for real $𝑎_{𝑘}$ (Proposition B.1.2(c)),

𝑝 (𝑧) = \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑧^{𝑘} = \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑧^{𝑘} = \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑧^{𝑘} = 𝑝 (𝑧) .

If $𝑝 (𝑧) = 0$ , then $𝑝 (𝑧) = 0 = 0$ . □

Example B.1.5 ( $𝑥^{4} + 4$ ). Let $𝑝 = 𝑥^{4} + 4$ . Since ${(1 + 𝑖)}^{2} = 2 𝑖$ , we get ${(1 + 𝑖)}^{4} = {(2 𝑖)}^{2} = - 4$ , so $𝑝 (1 + 𝑖) = 0$ . By Proposition B.1.4, $𝑝 (1 - 𝑖) = 0$ too. As $𝑝 (- 𝑧) = 𝑝 (𝑧)$ for every $𝑧$ , also $𝑝 (- 1 - 𝑖) =$ $𝑝 (- 1 + 𝑖) =$ $0$ . So $𝑝$ has the four roots $\pm 1 \pm 𝑖$ , and by Proposition 7.1.22 it has no others. (Example B.1.27 finds them again as the fourth roots of $- 4$ .)

The real coefficients matter: $𝑥 - 𝑖$ has the root $𝑖$ but not $- 𝑖$ .

Exercise B.1.6.routine Find all

𝑧 \in ℂ

with

𝑧^{2} = 𝑧^{2}

Solution.

Uses. Proposition B.1.2 (Conjugation respects the field operations), Proposition 1.1.17 (No zero divisors)

Hint 1. Move everything to one side and factor the difference of squares.

Hint 2.

(𝑧 - 𝑧) (𝑧 + 𝑧) = 0

. Now use Proposition B.1.2(c).

Sketch.

𝑧^{2} -

𝑧^{2} =

(𝑧 - 𝑧) (𝑧 + 𝑧)

, so the equation holds if and only if

𝑧 = 𝑧

𝑧 = - 𝑧

. The solutions are the real numbers and the purely imaginary numbers.

Since multiplication in

ℂ

is commutative,

(𝑧 - 𝑧) (𝑧 + 𝑧) =

𝑧^{2} -

𝑧^{2}

. So

𝑧^{2} = 𝑧^{2}

if and only if

(𝑧 - 𝑧) (𝑧 + 𝑧) = 0

, which by Proposition 1.1.17 happens if and only if

𝑧 = 𝑧

𝑧 = - 𝑧

. By Proposition B.1.2(c), the solutions are exactly the real numbers and the purely imaginary numbers, that is, the points of the two coordinate axes.

B.1.2The modulus

Definition B.1.7 (Modulus). The modulus (or absolute value) of

𝑧 = 𝑎 + 𝑏 𝑖 \in ℂ

is the real number

| 𝑧 | = \sqrt{𝑎^{2} + 𝑏^{2}}

So $| 𝑧 |$ is the length of the vector $(𝑎, 𝑏)$ , and $| 𝑧 -$ $𝑤 | =$ $\sqrt{{(𝑎 - 𝑐)}^{2} + {(𝑏 - 𝑑)}^{2}}$ is the distance between the points $𝑧$ and $𝑤$ of the plane. For a real number $𝑎 = 𝑎 + 0 𝑖$ , $| 𝑎 | = \sqrt{𝑎^{2}}$ is its usual absolute value, so the notation is consistent. For example, $| 3 - 4 𝑖 | = 5$ , $| 𝑖 | = 1$ , and $| cos 𝜃 + 𝑖 sin 𝜃 | = 1$ for every $𝜃 \in ℝ$ .

Proposition B.1.8 (Properties of the modulus). Let $𝑧, 𝑤 \in ℂ$ .

(a) $| 𝑧 | \geq 0$ , with equality if and only if $𝑧 = 0$ .
(b) ${| 𝑧 |}^{2} = 𝑧 𝑧$ , and $| 𝑧 | = | - 𝑧 | = | 𝑧 |$ .
(c)If $𝑧 \neq 0$ , then $𝑧^{- 1} = 𝑧 / {| 𝑧 |}^{2}$ .
(d) $| 𝑧 𝑤 | = | 𝑧 | | 𝑤 |$ , and if $𝑤 \neq 0$ , then $| 𝑧 / 𝑤 | = | 𝑧 | / | 𝑤 |$ .
(e) $| Re 𝑧 | \leq | 𝑧 |$ and $| Im 𝑧 | \leq | 𝑧 |$ . Moreover $Re 𝑧 \leq | 𝑧 |$ , with equality if and only if $𝑧$ is a nonnegative real number.
(f) ${| 𝑧 + 𝑤 |}^{2} =$ ${| 𝑧 |}^{2} +$ $2 Re (𝑧 𝑤) +$ ${| 𝑤 |}^{2}$ .

Most of these are one-line computations in coordinates. Parts (d) and (f) are better done without coordinates, by a move worth naming.

Technique B.1.9 (Square the modulus). The modulus is a square root, and square roots do not expand. Its square does: ${| 𝑧 |}^{2} = 𝑧 𝑧$ is a product, which can be multiplied out with the field axioms and conjugated term by term with Proposition B.1.2. Nothing is lost by squaring, since moduli are nonnegative: for real $𝑠, 𝑡 \geq 0$ , $𝑠 = 𝑡$ if and only if $𝑠^{2} = 𝑡^{2}$ , and $𝑠 \leq 𝑡$ if and only if $𝑠^{2} \leq 𝑡^{2}$ . So to prove an identity or an inequality between moduli, square both sides, write each ${| 𝑢 |}^{2}$ as $𝑢 𝑢$ , and expand.

Reach for it whenever moduli of sums or products appear. The canonical example is $| 𝑧 𝑤 | = | 𝑧 | | 𝑤 |$ in Proposition B.1.8, where ${| 𝑧 𝑤 |}^{2} =$ $𝑧 𝑤 𝑧 𝑤 =$ ${| 𝑧 |}^{2} {| 𝑤 |}^{2}$ . With ${‖ 𝐯 ‖}^{2} = ⟨ 𝐯, 𝐯 ⟩$ in place of ${| 𝑧 |}^{2} = 𝑧 𝑧$ , the same move expands norms in inner product spaces (Proposition 6.1.29).

Used in 6 places and 1 practice exercise

Proposition B.1.8, Theorem B.1.11, Exercise B.1.35, Exercise B.1.36, Exercise B.1.37, Exercise B.1.44; 1 practice exercise.

Proof.routine

Uses. Definition B.1.7 (Modulus), Proposition B.1.2 (Conjugation respects the field operations), Theorem 1.1.9 (Uniqueness of identities and inverses)

Techniques. Square the modulus (Technique B.1.9)

Hint 1. Parts (a), (b), and (e) are statements about

𝑎^{2} + 𝑏^{2}

for

𝑧 = 𝑎 + 𝑏 𝑖

. Compute in coordinates.

Hint 2. For (c), multiply

𝑧

by the proposed inverse and use (b). For (d), compare

{| 𝑧 𝑤 |}^{2}

with

{| 𝑧 |}^{2} {| 𝑤 |}^{2}

, writing each as a number times its conjugate.

Hint 3. For (f), expand

(𝑧 + 𝑤) (𝑧 + 𝑤)

and recognize the two cross terms as a number and its conjugate.

Sketch. Parts (a), (b), and (e) are coordinate facts about

𝑎^{2} + 𝑏^{2}

. Then

𝑧 \cdot 𝑧 / {| 𝑧 |}^{2} = 1

gives (c), and

{| 𝑧 𝑤 |}^{2} =

(𝑧 𝑤) (𝑧 𝑤) =

{| 𝑧 |}^{2} {| 𝑤 |}^{2}

gives (d). For (f),

(𝑧 + 𝑤) (𝑧 + 𝑤) =

{| 𝑧 |}^{2} +

(𝑢 + 𝑢) +

{| 𝑤 |}^{2}

with

𝑢 = 𝑧 𝑤

, and

𝑢 + 𝑢 = 2 Re 𝑢

Let $𝑧 = 𝑎 + 𝑏 𝑖$ .

(a) $𝑎^{2} + 𝑏^{2} \geq 0$ , so the square root is defined and nonnegative. It is $0$ if and only if $𝑎^{2} + 𝑏^{2} = 0$ , that is, $𝑎 = 𝑏 = 0$ .

(b) $𝑧 𝑧 =$ $(𝑎 + 𝑏 𝑖) (𝑎 - 𝑏 𝑖) =$ $(𝑎^{2} + 𝑏^{2}) +$ $(- 𝑎 𝑏 + 𝑏 𝑎) 𝑖 =$ $𝑎^{2} +$ $𝑏^{2} =$ ${| 𝑧 |}^{2}$ . Also $| 𝑧 | =$ $\sqrt{𝑎^{2} + {(- 𝑏)}^{2}} =$ $| 𝑧 |$ and $| -$ $𝑧 | =$ $\sqrt{{(- 𝑎)}^{2} + {(- 𝑏)}^{2}} =$ $| 𝑧 |$ .

(c) By (a), ${| 𝑧 |}^{2}$ is a nonzero real number, and by (b), $𝑧 \cdot$ $𝑧 / {| 𝑧 |}^{2} =$ ${| 𝑧 |}^{2} / {| 𝑧 |}^{2} =$ $1$ . So $𝑧 / {| 𝑧 |}^{2}$ is the inverse of $𝑧$ (Theorem 1.1.9).

(d) By (b) and Proposition B.1.2(a),

{| 𝑧 𝑤 |}^{2} = (𝑧 𝑤) 𝑧 𝑤 = (𝑧 𝑧) (𝑤 𝑤) = {| 𝑧 |}^{2} {| 𝑤 |}^{2} = {(| 𝑧 | | 𝑤 |)}^{2} .

Both $| 𝑧 𝑤 |$ and $| 𝑧 | | 𝑤 |$ are nonnegative, so they are equal. If $𝑤 \neq 0$ , then $| 𝑧 / 𝑤 | | 𝑤 | = | 𝑧 |$ by what we just proved, and dividing by $| 𝑤 | > 0$ gives $| 𝑧 / 𝑤 | = | 𝑧 | / | 𝑤 |$ .

(e) $| 𝑎 | =$ $\sqrt{𝑎^{2}} \leq$ $\sqrt{𝑎^{2} + 𝑏^{2}} =$ $| 𝑧 |$ , and likewise $| 𝑏 | \leq | 𝑧 |$ . Next, $Re 𝑧 = 𝑎 \leq | 𝑎 | \leq | 𝑧 |$ . The first inequality is an equality if and only if $𝑎 \geq 0$ , and the second if and only if $𝑏 = 0$ . So $Re 𝑧 = | 𝑧 |$ if and only if $𝑏 = 0$ and $𝑎 \geq 0$ , that is, $𝑧$ is a nonnegative real number.

(f) By (b) and Proposition B.1.2(a),

{| 𝑧 + 𝑤 |}^{2} = (𝑧 + 𝑤) (𝑧 + 𝑤) = 𝑧 𝑧 + (𝑧 𝑤 + 𝑧 𝑤) + 𝑤 𝑤 .

By Proposition B.1.2(a) and (b), $𝑧 𝑤 = 𝑧 𝑤$ , so the middle term is $𝑢 + 𝑢 = 2 Re 𝑢$ with $𝑢 = 𝑧 𝑤$ (Proposition B.1.2(d)). □

In coordinates, (d) says that $(𝑎^{2} + 𝑏^{2}) (𝑐^{2} + 𝑑^{2}) =$ ${(𝑎 𝑐 - 𝑏 𝑑)}^{2} +$ ${(𝑎 𝑑 + 𝑏 𝑐)}^{2}$ : a product of two sums of two squares is again a sum of two squares. Exercise B.1.48 puts this to use. Similarly, $Re (𝑧 𝑤) =$ $𝑎 𝑐 +$ $𝑏 𝑑$ is the dot product of $(𝑎, 𝑏)$ and $(𝑐, 𝑑)$ , so (f) is the law of cosines in the plane.

Exercise B.1.10.routine Let

𝑧 = (1 + 2 𝑖) / (2 - 𝑖)

. Find

| 𝑧 |

without computing

𝑧

. Then compute

𝑧

, and check your answer.

Solution.

Uses. Proposition B.1.8 (Properties of the modulus)

Hint 1. Use Proposition B.1.8(d) for the quotient.

Hint 2. To compute

𝑧

, multiply the numerator and the denominator by

2 - 𝑖 = 2 + 𝑖

; this is Proposition B.1.8(c).

Sketch.

| 𝑧 | =

| 1 +

2 𝑖 | / | 2 -

𝑖 | =

\sqrt{5} / \sqrt{5} =

1

. And

𝑧 =

(1 + 2 𝑖) (2 + 𝑖) / 5 =

5 𝑖 / 5 =

𝑖

, which has modulus

1

By Proposition B.1.8(d), $| 𝑧 | =$ $| 1 +$ $2 𝑖 | / | 2 -$ $𝑖 | =$ $\sqrt{1 + 4} / \sqrt{4 + 1} =$ $1$ . By Proposition B.1.8(c), ${(2 - 𝑖)}^{- 1} = (2 + 𝑖) / 5$ , so

𝑧 = \frac{(1 + 2 𝑖) (2 + 𝑖)}{5} = \frac{2 + 𝑖 + 4 𝑖 + 2 𝑖^{2}}{5} = \frac{5 𝑖}{5} = 𝑖,

and indeed $| 𝑖 | = 1$ .

The triangle inequality

The modulus is a length, so it should satisfy the triangle inequality. In the triangle with vertices $0$ , $𝑧$ , and $𝑧 + 𝑤$ , the sides have lengths $| 𝑧 |$ , $| 𝑤 |$ , and $| 𝑧 + 𝑤 |$ (Figure B.1.1(b)). With more terms, the broken path $0, 𝑧_{1}, 𝑧_{1} + 𝑧_{2}, \dots$ is at least as long as the straight segment from $0$ to its endpoint, and the two are equally long only if the path never turns.

Theorem B.1.11 (Triangle inequality in $ℂ$ ). Let $𝑛 \geq 1$ and $𝑧_{1}, \dots, 𝑧_{𝑛} \in ℂ$ .

(a) $| 𝑧_{1} +$ $\dots +$ $𝑧_{𝑛} | \leq$ $| 𝑧_{1} | +$ $\dots +$ $| 𝑧_{𝑛} |$ .
(b)Equality holds in (a) if and only if there is a $𝑢 \in ℂ$ with $| 𝑢 | = 1$ such that $𝑧_{𝑘} = | 𝑧_{𝑘} | 𝑢$ for every $𝑘$ . That is, all the $𝑧_{𝑘}$ point in the same direction $𝑢$ .

Intuition (Two ways in). For two terms, Technique B.1.9 applies. By Proposition B.1.8(f), ${| 𝑧 + 𝑤 |}^{2} =$ ${| 𝑧 |}^{2} +$ $2 Re (𝑧 𝑤) +$ ${| 𝑤 |}^{2}$ , while ${(| 𝑧 | + | 𝑤 |)}^{2} =$ ${| 𝑧 |}^{2} +$ $2 | 𝑧 𝑤 | +$ ${| 𝑤 |}^{2}$ . So the inequality says $Re (𝑧 𝑤) \leq | 𝑧 𝑤 |$ , which is Proposition B.1.8(e), and induction handles $𝑛$ terms.

Alternatively, rotate the whole picture so that the sum $𝑠$ points along the positive real axis (Technique 6.2.8). Then $| 𝑠 |$ is the real part of the rotated sum. The real part of a sum is the sum of the real parts, and each real part is at most the modulus of its term. The first argument is the one that generalizes to inner product spaces (Theorem 6.2.11). The second handles all $𝑛$ terms at once and makes the equality case transparent.

Proof (by squaring).standard

Uses. Proposition B.1.8 (Properties of the modulus), induction on $𝑛$

Techniques. Square the modulus (Technique B.1.9)

Hint 1. Prove the case

𝑛 = 2

first. Both sides are nonnegative, so compare squares, and expand

{| 𝑧 + 𝑤 |}^{2}

with Proposition B.1.8(f).

Hint 2. The case

𝑛 = 2

reduces to

Re (𝑧 𝑤) \leq | 𝑧 𝑤 |

, with equality exactly when

𝑧 𝑤

is a nonnegative real number. For

𝑛

terms, induct, splitting off

𝑧_{𝑛}

Hint 3. For the equality case with

𝑛

terms, equality forces equality in both steps of

| 𝑠 +

𝑧_{𝑛} | \leq

| 𝑠 | +

| 𝑧_{𝑛} | \leq

\sum_{𝑘} | 𝑧_{𝑘} |

, where

𝑠 = 𝑧_{1} + \dots + 𝑧_{𝑛 - 1}

. Treat

𝑠 = 0

separately.

Sketch. For two terms,

{| 𝑧 + 𝑤 |}^{2} =

{| 𝑧 |}^{2} +

2 Re (𝑧 𝑤) +

{| 𝑤 |}^{2} \leq

{(| 𝑧 | + | 𝑤 |)}^{2}

, with equality if and only if

𝑧 𝑤

is real and nonnegative. Induction on

𝑛

gives (a). For (b), “if” is a computation. For “only if”, equality for

𝑛

terms forces equality for

𝑧_{1}, \dots, 𝑧_{𝑛 - 1}

, so they point along some

𝑢

, and for the two terms

𝑠 =

𝑧_{1} +

\dots +

𝑧_{𝑛 - 1} =

| 𝑠 | 𝑢

and

𝑧_{𝑛}

. If

𝑠 \neq 0

, then

𝑢 𝑧_{𝑛} \geq 0

, which gives

𝑧_{𝑛} = | 𝑧_{𝑛} | 𝑢

; if

𝑠 = 0

, then

𝑧_{1} = \dots = 𝑧_{𝑛 - 1} = 0

, and the direction of

𝑧_{𝑛}

works for all terms.

Two terms. Let $𝑧, 𝑤 \in ℂ$ . By parts (f), (e), (d), and (b) of Proposition B.1.8,

\begin{aligned} {| 𝑧 + 𝑤 |}^{2} & = {| 𝑧 |}^{2} + 2 Re (𝑧 𝑤) + {| 𝑤 |}^{2} \\ \leq {| 𝑧 |}^{2} + 2 | 𝑧 𝑤 | + {| 𝑤 |}^{2} = {(| 𝑧 | + | 𝑤 |)}^{2}, \end{aligned}

with equality if and only if $Re (𝑧 𝑤) = | 𝑧 𝑤 |$ , that is (by Proposition B.1.8(e)), if and only if $𝑧 𝑤$ is a nonnegative real number. Both $| 𝑧 + 𝑤 |$ and $| 𝑧 | + | 𝑤 |$ are nonnegative, so

| 𝑧 + 𝑤 | \leq | 𝑧 | + | 𝑤 |, with equality if and only if 𝑧 𝑤 \in ℝ and 𝑧 𝑤 \geq 0 .

(B.1.1)

(a) We induct on $𝑛$ . For $𝑛 = 1$ there is nothing to prove. For $𝑛 \geq 2$ , let $𝑠 = 𝑧_{1} + \dots + 𝑧_{𝑛 - 1}$ . By (B.1.1) and the induction hypothesis,

| 𝑠 + 𝑧_{𝑛} | \leq | 𝑠 | + | 𝑧_{𝑛} | \leq | 𝑧_{1} | + \dots + | 𝑧_{𝑛 - 1} | + | 𝑧_{𝑛} | .

(B.1.2)

(b) Suppose $𝑧_{𝑘} = | 𝑧_{𝑘} | 𝑢$ for every $𝑘$ , with $| 𝑢 | = 1$ . Then $\sum_{𝑘} 𝑧_{𝑘} =$ $(\sum_{𝑘} | 𝑧_{𝑘} |) 𝑢$ , and by Proposition B.1.8(d) its modulus is $\sum_{𝑘} | 𝑧_{𝑘} |$ .

Conversely, suppose equality holds in (a). We find $𝑢$ by induction on $𝑛$ . For $𝑛 = 1$ , take $𝑢 = 𝑧_{1} / | 𝑧_{1} |$ if $𝑧_{1} \neq 0$ and $𝑢 = 1$ if $𝑧_{1} = 0$ . Let $𝑛 \geq 2$ . Equality between the two ends of (B.1.2) forces equality in both of its steps:

| 𝑠 + 𝑧_{𝑛} | = | 𝑠 | + | 𝑧_{𝑛} | and | 𝑠 | = | 𝑧_{1} | + \dots + | 𝑧_{𝑛 - 1} | .

By the induction hypothesis there is $𝑢$ with $| 𝑢 | = 1$ and $𝑧_{𝑘} = | 𝑧_{𝑘} | 𝑢$ for $𝑘 < 𝑛$ , and then $𝑠 = | 𝑠 | 𝑢$ by the computation in the previous paragraph.

If $𝑠 = 0$ , then $| 𝑧_{1} | +$ $\dots +$ $| 𝑧_{𝑛 - 1} | =$ $0$ , so $𝑧_{1} = \dots = 𝑧_{𝑛 - 1} = 0$ . Choose $𝑢^{'}$ for the single number $𝑧_{𝑛}$ as in the case $𝑛 = 1$ ; then $𝑧_{𝑘} = | 𝑧_{𝑘} | 𝑢^{'}$ for every $𝑘$ .
If $𝑠 \neq 0$ , then by the first equality and (B.1.1), $𝑠 𝑧_{𝑛} = | 𝑠 | 𝑢 𝑧_{𝑛}$ is a nonnegative real number. Dividing by $| 𝑠 | > 0$ , so is $𝑢 𝑧_{𝑛}$ , and hence $𝑢 𝑧_{𝑛} =$ $| 𝑢 𝑧_{𝑛} | =$ $| 𝑧_{𝑛} |$ . Conjugating gives $𝑢 𝑧_{𝑛} = | 𝑧_{𝑛} |$ , and multiplying by $𝑢$ , with $𝑢 𝑢 = {| 𝑢 |}^{2} = 1$ , gives $𝑧_{𝑛} = | 𝑧_{𝑛} | 𝑢$ .

□

Proof (by rotating the phase).standard

Uses. Proposition B.1.8 (Properties of the modulus), Definition 1.1.29 (Complex numbers)

Techniques. Rotate the phase (Technique 6.2.8)

Hint 1. Let

𝑠 = 𝑧_{1} + \dots + 𝑧_{𝑛}

. If

𝑠 \neq 0

, multiply

𝑠

𝑢

, where

𝑢 = 𝑠 / | 𝑠 |

. What is

𝑢 𝑠

Hint 2.

| 𝑠 | =

𝑢 𝑠 =

Re (𝑢 𝑠) =

\sum_{𝑘} Re (𝑢 𝑧_{𝑘})

. Bound each term with Proposition B.1.8(e).

Hint 3. Equality holds if and only if

Re (𝑢 𝑧_{𝑘}) =

| 𝑢 𝑧_{𝑘} |

for every

𝑘

. When does that happen?

Sketch. If

𝑠 = 0

, (a) is clear, and equality forces every

𝑧_{𝑘} = 0

. Otherwise let

𝑢 = 𝑠 / | 𝑠 |

. Then

| 𝑠 | =

Re (𝑢 𝑠) =

\sum_{𝑘} Re (𝑢 𝑧_{𝑘}) \leq

\sum_{𝑘} | 𝑢 𝑧_{𝑘} | =

\sum_{𝑘} | 𝑧_{𝑘} |

, with equality if and only if each

𝑢 𝑧_{𝑘}

is a nonnegative real number. As

| 𝑢 𝑧_{𝑘} | = | 𝑧_{𝑘} |

, that means

𝑢 𝑧_{𝑘} = | 𝑧_{𝑘} |

, that is,

𝑧_{𝑘} = | 𝑧_{𝑘} | 𝑢

If $𝑧_{𝑘} = | 𝑧_{𝑘} | 𝑢$ for all $𝑘$ with $| 𝑢 | = 1$ , then $\sum_{𝑘} 𝑧_{𝑘} =$ $(\sum_{𝑘} | 𝑧_{𝑘} |) 𝑢$ has modulus $\sum_{𝑘} | 𝑧_{𝑘} |$ by Proposition B.1.8(d). This is the “if” part of (b).

Let $𝑠 = 𝑧_{1} + \dots + 𝑧_{𝑛}$ . If $𝑠 = 0$ , then (a) holds because its right side is nonnegative, and equality means $| 𝑧_{1} | + \dots + | 𝑧_{𝑛} | = 0$ . Then every $𝑧_{𝑘}$ is $0$ , and $𝑢 = 1$ works.

Suppose $𝑠 \neq 0$ , and let $𝑢 = 𝑠 / | 𝑠 |$ . Then $| 𝑢 | = 1$ , and $𝑢 𝑠 =$ $𝑠 𝑠 / | 𝑠 | =$ $| 𝑠 |$ by Proposition B.1.8(b). The real part of a sum is the sum of the real parts, since addition in $ℂ$ is coordinatewise (Definition 1.1.29). With parts (e) and (d) of Proposition B.1.8,

| 𝑠 | = Re (𝑢 𝑠) = \sum_{𝑘 = 1}^{𝑛} Re (𝑢 𝑧_{𝑘}) \leq \sum_{𝑘 = 1}^{𝑛} | 𝑢 𝑧_{𝑘} | = \sum_{𝑘 = 1}^{𝑛} | 𝑧_{𝑘} | .

This is (a). If equality holds, then $Re (𝑢 𝑧_{𝑘}) =$ $| 𝑢 𝑧_{𝑘} |$ for every $𝑘$ , because a sum of nonnegative gaps vanishes only if each gap does. By Proposition B.1.8(e), each $𝑢 𝑧_{𝑘}$ is then a nonnegative real number, and since $| 𝑢 𝑧_{𝑘} | = | 𝑧_{𝑘} |$ , this says $𝑢 𝑧_{𝑘} = | 𝑧_{𝑘} |$ . Multiplying by $𝑢$ gives $𝑧_{𝑘} = | 𝑧_{𝑘} | 𝑢$ for every $𝑘$ . □

The first proof found the direction $𝑢$ one term at a time; the second named it at the start, as the direction of the sum, and then checked each term against it.

Corollary B.1.12 (Two terms, and the reverse triangle inequality). Let $𝑧, 𝑤 \in ℂ$ .

(a) $| 𝑧 + 𝑤 | = | 𝑧 | + | 𝑤 |$ if and only if one of $𝑧, 𝑤$ is a nonnegative real multiple of the other, if and only if $𝑧 𝑤$ is a nonnegative real number.
(b) $| | 𝑧 | -$ $| 𝑤 | | \leq$ $| 𝑧 -$ $𝑤 |$ .

Proof.routine

Uses. Theorem B.1.11 (Triangle inequality in $ℂ$ ), Proposition B.1.8 (Properties of the modulus)

Hint 1. For (a), take

𝑛 = 2

in Theorem B.1.11(b), and translate “

𝑧 = | 𝑧 | 𝑢

and

𝑤 = | 𝑤 | 𝑢

” into the other two conditions.

Hint 2. For (b), write

𝑧 = (𝑧 - 𝑤) + 𝑤

and apply the triangle inequality; then exchange the roles of

𝑧

and

𝑤

Sketch. (a) If

𝑧 = | 𝑧 | 𝑢

and

𝑤 = | 𝑤 | 𝑢

, then

𝑧 𝑤 = | 𝑧 | | 𝑤 | \geq 0

. If

𝑧 𝑤 = 𝑡 \geq 0

and

𝑤 \neq 0

, then

𝑧 = (𝑡 / {| 𝑤 |}^{2}) 𝑤

. If

𝑧 = 𝑡 𝑤

with

𝑡 \geq 0

and

𝑤 \neq 0

, then

𝑢 = 𝑤 / | 𝑤 |

works. (b)

| 𝑧 | \leq | 𝑧 - 𝑤 | + | 𝑤 |

and

| 𝑤 | \leq | 𝑤 - 𝑧 | + | 𝑧 |

(a) By Theorem B.1.11(b) with $𝑛 = 2$ , the equality holds if and only if (i) $𝑧 = | 𝑧 | 𝑢$ and $𝑤 = | 𝑤 | 𝑢$ for some $𝑢$ with $| 𝑢 | = 1$ . Let (ii) say that $𝑧 𝑤$ is a nonnegative real number, and (iii) that one of $𝑧, 𝑤$ is a nonnegative real multiple of the other. We prove (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i).

If (i) holds, then $𝑧 𝑤 =$ $| 𝑧 | | 𝑤 | 𝑢 𝑢 =$ $| 𝑧 | | 𝑤 | \geq$ $0$ .

Suppose $𝑧 𝑤 = 𝑡$ with $𝑡$ real and $𝑡 \geq 0$ . If $𝑤 = 0$ , then $𝑤 = 0 \cdot 𝑧$ . If $𝑤 \neq 0$ , multiplying by $𝑤$ gives $𝑧 {| 𝑤 |}^{2} = 𝑡 𝑤$ , so $𝑧 = (𝑡 / {| 𝑤 |}^{2}) 𝑤$ with $𝑡 / {| 𝑤 |}^{2} \geq 0$ .

Suppose finally that $𝑧 = 𝑡 𝑤$ with $𝑡 \geq 0$ ; the case $𝑤 = 𝑡 𝑧$ is the same with $𝑧$ and $𝑤$ exchanged, and (i) is symmetric in $𝑧$ and $𝑤$ . If $𝑤 = 0$ , then $𝑧 = 0$ as well, and $𝑢 = 1$ works. If $𝑤 \neq 0$ , let $𝑢 = 𝑤 / | 𝑤 |$ . Then $𝑤 = | 𝑤 | 𝑢$ and $𝑧 = 𝑡 | 𝑤 | 𝑢$ , where $𝑡 | 𝑤 | =$ $| 𝑡 𝑤 | =$ $| 𝑧 |$ by Proposition B.1.8(d).

(b) By Theorem B.1.11, $| 𝑧 | =$ $| (𝑧 - 𝑤) +$ $𝑤 | \leq$ $| 𝑧 -$ $𝑤 | +$ $| 𝑤 |$ , so $| 𝑧 | - | 𝑤 | \leq | 𝑧 - 𝑤 |$ . Exchanging $𝑧$ and $𝑤$ , and using $| 𝑤 -$ $𝑧 | =$ $| -$ $(𝑧 - 𝑤) | =$ $| 𝑧 -$ $𝑤 |$ (Proposition B.1.8(b)), gives $| 𝑤 | - | 𝑧 | \leq | 𝑧 - 𝑤 |$ . Together these say $| | 𝑧 | -$ $| 𝑤 | | \leq$ $| 𝑧 -$ $𝑤 |$ . □

With the dot product $𝑧 𝑤$ on $ℂ = ℂ^{1}$ (Definition 6.1.8), $| 𝑧 |$ is the norm, and Corollary B.1.12(a) is the case $𝑉 = ℂ$ of Theorem 6.2.11(b). The proofs above use only Proposition B.1.8, so Chapter 6 can rely on them.

B.1.3The plane and polar form

As a real vector space, $ℂ$ is $ℝ^{2}$ with the basis $1, 𝑖$ (Example 1.6.5). So adding complex numbers is adding vectors, by the parallelogram rule, and multiplying by a real number scales (Figure B.1.1). What is new is multiplication by a nonreal number, and to describe it we need polar coordinates.

Figure B.1.1. (a) The number

𝑧 = 𝑎 + 𝑏 𝑖

as the point

(𝑎, 𝑏)

, with its modulus

| 𝑧 |

and an argument

𝜃

, and its conjugate

𝑧

, the mirror image in the real axis. (b) Addition is vector addition. The triangle

0, 𝑧, 𝑧 + 𝑤

has sides of lengths

| 𝑧 |

| 𝑤 |

, and

| 𝑧 + 𝑤 |

, and the red diagonal of the parallelogram has length

| 𝑧 - 𝑤 |

Remark B.1.13 (Facts about

cos

and

sin

). We use the following facts about the real functions

cos

and

sin

, from calculus.

(T1) ${cos}^{2} 𝜃 + {sin}^{2} 𝜃 = 1$ , $cos (- 𝜃) = cos 𝜃$ , $sin (- 𝜃) = - sin 𝜃$ , and the angle addition formulas
$cos (𝛼 + 𝛽) = cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽, sin (𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 .$
(T2) $cos 𝜃 = cos 𝜃^{'}$ and $sin 𝜃 = sin 𝜃^{'}$ if and only if $𝜃 -$ $𝜃^{'} \in$ $2 𝜋 ℤ =$ ${2 𝜋 𝑘 | 𝑘 \in ℤ}$ .
(T3)For every $(𝑥, 𝑦) \in ℝ^{2}$ with $𝑥^{2} + 𝑦^{2} = 1$ there is a $𝜃 \in ℝ$ with $(𝑥, 𝑦) =$ $(cos 𝜃, sin 𝜃)$ .

We used (T1) to show that rotations are linear (Example 2.1.11).

Definition B.1.14 (Polar form and argument). Let $𝑧 \in ℂ$ be nonzero. An argument of $𝑧$ is a real number $𝜃$ with

𝑧 = | 𝑧 | (cos 𝜃 + 𝑖 sin 𝜃) .

Writing $𝑧 =$ $𝑟 (cos 𝜃 + 𝑖 sin 𝜃)$ with $𝑟 > 0$ and $𝜃 \in ℝ$ is writing $𝑧$ in polar form.

Proposition B.1.15 (Existence and uniqueness of polar form). Let $𝑧 \in ℂ$ be nonzero.

(a) $𝑧$ has an argument.
(b)If $𝑧 =$ $𝑟 (cos 𝜃 + 𝑖 sin 𝜃)$ with $𝑟 > 0$ and $𝜃 \in ℝ$ , then $𝑟 = | 𝑧 |$ , so $𝜃$ is an argument of $𝑧$ .
(c)If $𝜃$ is an argument of $𝑧$ , then the arguments of $𝑧$ are exactly the numbers $𝜃 + 2 𝜋 𝑘$ with $𝑘 \in ℤ$ .

Proof.routine

Uses. Definition B.1.14 (Polar form and argument), Proposition B.1.8 (Properties of the modulus), Remark B.1.13 (Facts about $cos$ and $sin$ )

Hint 1. For (a), the number

𝑧 / | 𝑧 |

has modulus

1

, so its coordinates

(𝑥, 𝑦)

satisfy

𝑥^{2} + 𝑦^{2} = 1

. Apply (T3).

Hint 2. For (b), take moduli of both sides. For (c), divide by

| 𝑧 |

and compare coordinates using (T2).

Sketch. (a)

𝑧 / | 𝑧 | = 𝑥 + 𝑦 𝑖

with

𝑥^{2} + 𝑦^{2} = 1

, and (T3) writes

(𝑥, 𝑦) =

(cos 𝜃, sin 𝜃)

. (b) Taking moduli,

| 𝑧 | =

𝑟 | cos 𝜃 +

𝑖 sin 𝜃 | =

𝑟

. (c)

𝜃^{'}

is an argument if and only if

cos 𝜃^{'} +

𝑖 sin 𝜃^{'} =

𝑧 / | 𝑧 | =

cos 𝜃 +

𝑖 sin 𝜃

, if and only if

𝜃^{'} - 𝜃 \in 2 𝜋 ℤ

by (T2).

(a) Let $𝑧 / | 𝑧 | = 𝑥 + 𝑦 𝑖$ . By Proposition B.1.8(d), $𝑥^{2} +$ $𝑦^{2} =$ ${| 𝑧 / | 𝑧 | |}^{2} =$ $1$ , so by (T3) there is a $𝜃$ with $𝑥 = cos 𝜃$ and $𝑦 = sin 𝜃$ . Then $𝑧 =$ $| 𝑧 | (cos 𝜃 + 𝑖 sin 𝜃)$ .

(b) By Proposition B.1.8(d) and (T1), $| 𝑧 | =$ $𝑟 | cos 𝜃 +$ $𝑖 sin 𝜃 | =$ $𝑟 \sqrt{{cos}^{2} 𝜃 + {sin}^{2} 𝜃} =$ $𝑟$ .

(c) Dividing by $| 𝑧 |$ , a real number $𝜃^{'}$ is an argument of $𝑧$ if and only if $cos 𝜃^{'} +$ $𝑖 sin 𝜃^{'} =$ $𝑧 / | 𝑧 | =$ $cos 𝜃 +$ $𝑖 sin 𝜃$ . Comparing real and imaginary parts, this holds if and only if $cos 𝜃^{'} = cos 𝜃$ and $sin 𝜃^{'} = sin 𝜃$ , which by (T2) holds if and only if $𝜃^{'} - 𝜃 \in 2 𝜋 ℤ$ . □

So two nonzero complex numbers are equal if and only if they have the same modulus and their arguments differ by a multiple of $2 𝜋$ . The number $0$ has no argument.

Example B.1.16 (Polar forms).

1 +

𝑖 =

\sqrt{2} (cos \frac{𝜋}{4} + 𝑖 sin \frac{𝜋}{4})

- 2 =

2 (cos 𝜋 + 𝑖 sin 𝜋)

- 3 𝑖 =

3 (cos (- \frac{𝜋}{2}) + 𝑖 sin (- \frac{𝜋}{2}))

, and

\sqrt{3} -

𝑖 =

2 (cos (- \frac{𝜋}{6}) + 𝑖 sin (- \frac{𝜋}{6}))

. Any argument may be changed by a multiple of

2 𝜋

: also

- 3 𝑖 =

3 (cos \frac{3 𝜋}{2} + 𝑖 sin \frac{3 𝜋}{2})

Proposition B.1.17 (Multiplication in polar form). For all $𝛼, 𝛽 \in ℝ$ ,

(cos 𝛼 + 𝑖 sin 𝛼) (cos 𝛽 + 𝑖 sin 𝛽) = cos (𝛼 + 𝛽) + 𝑖 sin (𝛼 + 𝛽) .

Consequently, if $𝑧 =$ $𝑟 (cos 𝛼 + 𝑖 sin 𝛼)$ and $𝑤 =$ $𝑠 (cos 𝛽 + 𝑖 sin 𝛽)$ with $𝑟, 𝑠 > 0$ , then

\begin{aligned} 𝑧 𝑤 & = 𝑟 𝑠 (cos (𝛼 + 𝛽) + 𝑖 sin (𝛼 + 𝛽)), \\ 𝑧^{- 1} & = 𝑟^{- 1} (cos (- 𝛼) + 𝑖 sin (- 𝛼)), \\ 𝑧 & = 𝑟 (cos (- 𝛼) + 𝑖 sin (- 𝛼)) . \end{aligned}

That is, moduli multiply and arguments add, and inverting or conjugating negates the argument.

Proof.routine

Uses. Definition 1.1.29 (Complex numbers), Remark B.1.13 (Facts about $cos$ and $sin$ )

Hint 1. Multiply out the left side with

𝑖^{2} = - 1

and compare with the angle addition formulas (T1).

Hint 2. For

𝑧^{- 1}

, multiply

𝑧

by the proposed inverse, using the first part with

𝛽 = - 𝛼

Sketch. Expanding gives

(cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽) +

(sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽) 𝑖

, which is the right side by (T1). The formula for

𝑧 𝑤

follows by regrouping. The proposed inverse times

𝑧

cos 0 + 𝑖 sin 0 = 1

, and

𝑧 =

𝑟 (cos 𝛼 - 𝑖 sin 𝛼)

is the stated number by (T1).

Expanding with $𝑖^{2} = - 1$ ,

(cos 𝛼 + 𝑖 sin 𝛼) (cos 𝛽 + 𝑖 sin 𝛽) = (cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽) + (sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽) 𝑖,

which is $cos (𝛼 + 𝛽) +$ $𝑖 sin (𝛼 + 𝛽)$ by (T1). For $𝑧$ and $𝑤$ as stated, $𝑧 𝑤 =$ $𝑟 𝑠 (cos 𝛼 + 𝑖 sin 𝛼) (cos 𝛽 + 𝑖 sin 𝛽)$ by commutativity and associativity, which gives the formula for $𝑧 𝑤$ . With $𝛽 = - 𝛼$ , the first part gives

𝑧 \cdot 𝑟^{- 1} (cos (- 𝛼) + 𝑖 sin (- 𝛼)) = cos 0 + 𝑖 sin 0 = 1,

which is the formula for $𝑧^{- 1}$ . Finally $𝑧 =$ $𝑟 (cos 𝛼 - 𝑖 sin 𝛼) =$ $𝑟 (cos (- 𝛼) + 𝑖 sin (- 𝛼))$ by (T1). □

Figure B.1.2. Multiplying

𝑧 =

| 𝑧 | (cos 𝛼 + 𝑖 sin 𝛼)

𝑤 =

| 𝑤 | (cos 𝛽 + 𝑖 sin 𝛽)

. The product

𝑧 𝑤

has modulus

| 𝑧 | | 𝑤 |

and argument

𝛼 + 𝛽

. The faint arc is part of the unit circle.

The computation in the proof is the one that shows that rotating by $𝛽$ and then by $𝛼$ is rotating by $𝛼 + 𝛽$ (Exercise 2.1.43). The next proposition explains why.

B.1.4Multiplication as rotation and scaling

Proposition B.1.18 (Multiplication by $𝑤$ as a map of the plane). Let $𝑤 = 𝑐 + 𝑑 𝑖 \in ℂ$ , and let $𝑀_{𝑤} : ℂ \to ℂ$ , $𝑀_{𝑤} (𝑧) = 𝑤 𝑧$ . Regard $ℂ$ as a vector space over $ℝ$ with the basis $ℬ︀ = 1, 𝑖$ .

(a) $𝑀_{𝑤}$ is $ℝ$ -linear, and
${[𝑀_{𝑤}]}_{ℬ︀ ℬ︀} = (\begin{matrix} 𝑐 & - 𝑑 \\ 𝑑 & 𝑐 \end{matrix}), where 𝑐^{2} + 𝑑^{2} = {| 𝑤 |}^{2} .$
(b)If $𝑤 \neq 0$ and $𝜃$ is an argument of $𝑤$ , then
${[𝑀_{𝑤}]}_{ℬ︀ ℬ︀} = | 𝑤 | (\begin{matrix} cos 𝜃 & - sin 𝜃 \\ sin 𝜃 & cos 𝜃 \end{matrix}) = | 𝑤 | {[𝑅_{𝜃}]}_{ℰ︀ ℰ︀},$
where $𝑅_{𝜃}$ is the rotation of $ℝ^{2}$ by $𝜃$ (Example 2.1.11). So, identifying $𝑎 + 𝑏 𝑖$ with $(𝑎, 𝑏)$ , multiplication by $𝑤$ is the rotation by $𝜃$ followed by scaling by $| 𝑤 |$ .

Proof.routine

Uses. Exercise 2.1.5, Definition 2.3.1 (Matrix of a linear map), Example 2.3.3 (Matrices with respect to standard bases), Definition B.1.14 (Polar form and argument)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. Linearity comes from distributivity and commutativity in

ℂ

. The columns of

{[𝑀_{𝑤}]}_{ℬ︀ ℬ︀}

are the coordinates of

𝑀_{𝑤} (1)

and

𝑀_{𝑤} (𝑖)

Hint 2. For (b), substitute

𝑐 = | 𝑤 | cos 𝜃

and

𝑑 = | 𝑤 | sin 𝜃

, and recall the matrix of

𝑅_{𝜃}

from Example 2.3.3.

Sketch.

𝑀_{𝑤}

is even

ℂ

-linear (Exercise 2.1.5), hence

ℝ

-linear.

𝑀_{𝑤} (1) = 𝑐 + 𝑑 𝑖

and

𝑀_{𝑤} (𝑖) = - 𝑑 + 𝑐 𝑖

give the columns

(𝑐, 𝑑)

and

(- 𝑑, 𝑐)

. In polar form

𝑐 = | 𝑤 | cos 𝜃

and

𝑑 = | 𝑤 | sin 𝜃

, and the matrix becomes

| 𝑤 |

times the rotation matrix.

(a) By Exercise 2.1.5 with $𝐹 = ℂ$ , $𝑀_{𝑤}$ is $ℂ$ -linear: $𝑤 (𝑧 + 𝑧^{'}) =$ $𝑤 𝑧 +$ $𝑤 𝑧^{'}$ and $𝑤 (𝜆 𝑧) = 𝜆 (𝑤 𝑧)$ for $𝜆 \in ℂ$ . In particular this holds for real $𝜆$ , so $𝑀_{𝑤}$ is $ℝ$ -linear. Its values on $ℬ︀$ are

𝑀_{𝑤} (1) = 𝑐 + 𝑑 𝑖 = 𝑐 \cdot 1 + 𝑑 \cdot 𝑖, 𝑀_{𝑤} (𝑖) = 𝑐 𝑖 + 𝑑 𝑖^{2} = (- 𝑑) \cdot 1 + 𝑐 \cdot 𝑖,

so by Definition 2.3.1 the columns of ${[𝑀_{𝑤}]}_{ℬ︀ ℬ︀}$ are $(𝑐, 𝑑)$ and $(- 𝑑, 𝑐)$ .

(b) By Definition B.1.14, $𝑐 +$ $𝑑 𝑖 =$ $| 𝑤 | (cos 𝜃 + 𝑖 sin 𝜃)$ , so $𝑐 = | 𝑤 | cos 𝜃$ and $𝑑 = | 𝑤 | sin 𝜃$ . Substituting into (a) gives the first equality, and the second is the matrix of $𝑅_{𝜃}$ from Example 2.3.3. The coordinate map $𝑎 + 𝑏 𝑖 \mapsto (𝑎, 𝑏)$ sends the basis $ℬ︀$ to the standard basis $ℰ︀$ of $ℝ^{2}$ , so under it $𝑀_{𝑤}$ becomes the linear map of $ℝ^{2}$ with standard matrix $| 𝑤 | {[𝑅_{𝜃}]}_{ℰ︀ ℰ︀}$ , namely $| 𝑤 | 𝑅_{𝜃}$ . □

Part (a) is also part (a) of Exercise 2.3.35. The determinant of the matrix is $𝑐^{2} + 𝑑^{2} = {| 𝑤 |}^{2}$ (Definition 5.1.14): multiplication by $𝑤$ scales areas by ${| 𝑤 |}^{2}$ , as a rotation followed by scaling every direction by $| 𝑤 |$ should. The shape of the matrix records $𝑀_{𝑤} (𝑖) =$ $𝑖 𝑀_{𝑤} (1)$ : its second column is its first, turned a quarter turn counterclockwise. An $ℝ$ -linear map of the plane has this property exactly when it is $ℂ$ -linear, that is, multiplication by a complex number (Exercise 2.1.47), and exactly when it commutes with the quarter turn $𝑅_{𝜋 / 2}$ (Exercise 2.2.30).

In the plane transformer below, $𝐴 = {[𝑀_{𝑤}]}_{ℬ︀ ℬ︀}$ for $𝑤 = 1 + 𝑖$ . So $𝐴 𝐞_{1} = (1, 1)$ is $𝑤$ and $𝐴 𝐞_{2} = (- 1, 1)$ is $𝑖 𝑤$ . The grid turns by $𝜋 / 4$ and stretches by $| 𝑤 | = \sqrt{2}$ , and $det 𝐴 = 2 = {| 𝑤 |}^{2}$ . Drag $𝐴 𝐞_{1}$ to another $𝑤$ , and then $𝐴 𝐞_{2}$ to its quarter turn $𝑖 𝑤$ : the grid stays square, and the determinant is ${| 𝑤 |}^{2}$ . In any other position of $𝐴 𝐞_{2}$ the map is still $ℝ$ -linear, but it is not multiplication by a complex number. Squares become non-square parallelograms, except at $𝐴 𝐞_{2} = - 𝑖 𝑤$ , where the map is $𝑧 \mapsto 𝑤 𝑧$ : it keeps squares square but reverses orientation, and $det 𝐴 = - {| 𝑤 |}^{2}$ .

Multiplication by

𝑤 = 1 + 𝑖

as a map of

ℂ = ℝ^{2}

. The unit square turns by

𝜋 / 4

and grows by the factor

| 𝑤 | = \sqrt{2}

in each direction, so its area becomes

{| 𝑤 |}^{2} = det 𝐴 = 2

The notation $𝑒^{𝑖 𝜃}$

Notation B.1.19 ( $𝑒^{𝑖 𝜃}$ ). For $𝜃 \in ℝ$ we write

𝑒^{𝑖 𝜃} = cos 𝜃 + 𝑖 sin 𝜃 .

In this notation a nonzero $𝑧$ with argument $𝜃$ is $𝑧 = | 𝑧 | 𝑒^{𝑖 𝜃}$ , and Proposition B.1.17, Proposition B.1.8, and (T2) say that

\begin{matrix} 𝑒^{𝑖 𝛼} 𝑒^{𝑖 𝛽} = 𝑒^{𝑖 (𝛼 + 𝛽)}, 𝑒^{𝑖 𝜃} = 𝑒^{- 𝑖 𝜃} = {(𝑒^{𝑖 𝜃})}^{- 1}, | 𝑒^{𝑖 𝜃} | = 1, \\ 𝑒^{𝑖 𝜃} = 1 if and only if 𝜃 \in 2 𝜋 ℤ . \end{matrix}

For example, $𝑒^{𝑖 𝜋 / 2} = 𝑖$ , $𝑒^{𝑖 𝜋} = - 1$ , and $𝑒^{2 𝜋 𝑖} = 1$ . For us $𝑒^{𝑖 𝜃}$ is an abbreviation, chosen because the first identity is a law of exponents. It is also the value of a genuine exponential function, but we never need that.

Aside (Why an exponential).Beyond 21-242 The power series

exp (𝑧) =

\sum_{𝑛 \geq 0} 𝑧^{𝑛} / 𝑛!

converges for every

𝑧 \in ℂ

, and multiplying two such series, with the binomial theorem, gives

exp (𝑧 + 𝑤) =

exp (𝑧) exp (𝑤)

. Splitting

exp (𝑖 𝜃)

into its even and odd terms and using

𝑖^{2 𝑘} = {(- 1)}^{𝑘}

gives the Taylor series of

cos 𝜃

and

𝑖 sin 𝜃

. So

exp (𝑖 𝜃) =

cos 𝜃 +

𝑖 sin 𝜃

(Euler’s formula), and

exp (𝑎 + 𝑖 𝑏) =

𝑒^{𝑎} 𝑒^{𝑖 𝑏}

for real

𝑎, 𝑏

. All of this needs convergence of series, which belongs to analysis.

Powers

Theorem B.1.20 (De Moivre’s formula). For every $𝜃 \in ℝ$ and every integer $𝑛$ ,

{(cos 𝜃 + 𝑖 sin 𝜃)}^{𝑛} = cos 𝑛 𝜃 + 𝑖 sin 𝑛 𝜃, that is, {(𝑒^{𝑖 𝜃})}^{𝑛} = 𝑒^{𝑖 𝑛 𝜃} .

Consequently ${(𝑟 𝑒^{𝑖 𝜃})}^{𝑛} =$ $𝑟^{𝑛} 𝑒^{𝑖 𝑛 𝜃}$ for $𝑟 > 0$ .

Proof.routine

Uses. Proposition B.1.17 (Multiplication in polar form), Notation B.1.19 ( $𝑒^{𝑖 𝜃}$ ), induction on $𝑛$

Hint 1. For

𝑛 \geq 0

, induct on

𝑛

, using

𝑒^{𝑖 (𝑛 + 1) 𝜃} =

𝑒^{𝑖 𝑛 𝜃} 𝑒^{𝑖 𝜃}

Hint 2. For

𝑛 < 0

, write

{(𝑒^{𝑖 𝜃})}^{𝑛} =

{({(𝑒^{𝑖 𝜃})}^{- 1})}^{- 𝑛}

and use

{(𝑒^{𝑖 𝜃})}^{- 1} =

𝑒^{𝑖 (- 𝜃)}

Sketch. The case

𝑛 = 0

1 = 1

, and

𝑒^{𝑖 𝑛 𝜃} 𝑒^{𝑖 𝜃} =

𝑒^{𝑖 (𝑛 + 1) 𝜃}

is the induction step. For negative

𝑛

, apply the nonnegative case to

- 𝜃

and

- 𝑛

, since

{(𝑒^{𝑖 𝜃})}^{- 1} =

𝑒^{- 𝑖 𝜃}

For $𝑛 = 0$ both sides are $1$ . If the formula holds for some $𝑛 \geq 0$ , then by Proposition B.1.17

{(𝑒^{𝑖 𝜃})}^{𝑛 + 1} = {(𝑒^{𝑖 𝜃})}^{𝑛} 𝑒^{𝑖 𝜃} = 𝑒^{𝑖 𝑛 𝜃} 𝑒^{𝑖 𝜃} = 𝑒^{𝑖 (𝑛 + 1) 𝜃} .

So it holds for every $𝑛 \geq 0$ . If $𝑛 < 0$ , then $- 𝑛 > 0$ , and since ${(𝑒^{𝑖 𝜃})}^{- 1} =$ $𝑒^{𝑖 (- 𝜃)}$ (Notation B.1.19), the case already proved, applied to $- 𝜃$ and $- 𝑛$ , gives

{(𝑒^{𝑖 𝜃})}^{𝑛} = {({(𝑒^{𝑖 𝜃})}^{- 1})}^{- 𝑛} = {(𝑒^{𝑖 (- 𝜃)})}^{- 𝑛} = 𝑒^{𝑖 (- 𝜃) (- 𝑛)} = 𝑒^{𝑖 𝑛 𝜃} .

Finally, ${(𝑟 𝑒^{𝑖 𝜃})}^{𝑛} =$ $𝑟^{𝑛} {(𝑒^{𝑖 𝜃})}^{𝑛} =$ $𝑟^{𝑛} 𝑒^{𝑖 𝑛 𝜃}$ . □

Technique B.1.21 (Use polar form for products, powers, and roots). Addition is easy in the form $𝑎 + 𝑏 𝑖$ . Multiplication is easy in polar form, where moduli multiply and arguments add (Proposition B.1.17). So when a problem is about products, quotients, powers, or roots of complex numbers, or about rotations of the plane, write the numbers as $𝑟 𝑒^{𝑖 𝜃}$ . The problem then splits into a question about positive real numbers (the moduli) and a question about angles modulo $2 𝜋$ (the arguments), and each is usually easy. Converting back to the form $𝑎 + 𝑏 𝑖$ costs a cosine and a sine.

Reach for it to compute a high power, to solve $𝑧^{𝑛} = 𝑤$ , or to see what a product does geometrically. The canonical example is the solution of $𝑧^{𝑛} = 𝑤$ in Theorem B.1.26.

Used in 6 places and 2 practice exercises

Exercise B.1.23, Theorem B.1.25, Theorem B.1.26, Proposition B.1.29, Exercise B.1.33, Exercise B.1.34; 2 practice exercises.

Example B.1.22 (Powers and multiple angles).

(a) $1 + 𝑖 = \sqrt{2} 𝑒^{𝑖 𝜋 / 4}$ , so by Theorem B.1.20, ${(1 + 𝑖)}^{8} =$ ${(\sqrt{2})}^{8} 𝑒^{2 𝜋 𝑖} =$ $16$ . Compare the repeated squaring in Exercise 1.1.36(d).
(b) $\sqrt{3} - 𝑖 = 2 𝑒^{- 𝑖 𝜋 / 6}$ , so ${(\sqrt{3} - 𝑖)}^{- 6} =$ $2^{- 6} 𝑒^{𝑖 𝜋} =$ $- 1 / 64$ .
(c)Expand the left side of ${(cos 𝜃 + 𝑖 sin 𝜃)}^{3} =$ $cos 3 𝜃 +$ $𝑖 sin 3 𝜃$ with the binomial theorem and compare real parts:
$cos 3 𝜃 = {cos}^{3} 𝜃 - 3 cos 𝜃 {sin}^{2} 𝜃 = 4 {cos}^{3} 𝜃 - 3 cos 𝜃,$
using ${sin}^{2} 𝜃 = 1 - {cos}^{2} 𝜃$ . The imaginary parts give $sin 3 𝜃 =$ $3 {cos}^{2} 𝜃 sin 𝜃 -$ ${sin}^{3} 𝜃 =$ $3 sin 𝜃 -$ $4 {sin}^{3} 𝜃$ .

Exercise B.1.23.routine Write

- 2 + 2 𝑖

and

- 1 - \sqrt{3} 𝑖

in polar form, and compute

{(- 2 + 2 𝑖)}^{5} / {(- 1 - \sqrt{3} 𝑖)}^{3}

Solution.

Uses. Proposition B.1.15 (Existence and uniqueness of polar form), Theorem B.1.20 (De Moivre’s formula), Proposition B.1.17 (Multiplication in polar form)

Techniques. Use polar form for products, powers, and roots (Technique B.1.21)

Hint 1. The moduli are

2 \sqrt{2}

and

2

. Find angles with the right cosine and sine.

Hint 2.

- 2 +

2 𝑖 =

2 \sqrt{2} 𝑒^{3 𝜋 𝑖 / 4}

and

- 1 -

\sqrt{3} 𝑖 =

2 𝑒^{- 2 𝜋 𝑖 / 3}

. Now apply Theorem B.1.20 and divide.

Sketch.

{(- 2 + 2 𝑖)}^{5} =

{(2 \sqrt{2})}^{5} 𝑒^{15 𝜋 𝑖 / 4} =

128 \sqrt{2} 𝑒^{- 𝑖 𝜋 / 4} =

128 -

128 𝑖

and

{(- 1 - \sqrt{3} 𝑖)}^{3} =

8 𝑒^{- 2 𝜋 𝑖} =

8

, so the quotient is

16 - 16 𝑖

We have $| - 2 + 2 𝑖 | = \sqrt{8} = 2 \sqrt{2}$ and $(- 2 + 2 𝑖) / (2 \sqrt{2}) =$ $- 1 / \sqrt{2} +$ $𝑖 / \sqrt{2} =$ $𝑒^{3 𝜋 𝑖 / 4}$ , so $- 2 +$ $2 𝑖 =$ $2 \sqrt{2} 𝑒^{3 𝜋 𝑖 / 4}$ . Similarly $| - 1 - \sqrt{3} 𝑖 | = 2$ and $- 1 -$ $\sqrt{3} 𝑖 =$ $2 (- 1 / 2 - (\sqrt{3} / 2) 𝑖) =$ $2 𝑒^{- 2 𝜋 𝑖 / 3}$ . By Theorem B.1.20,

{(- 2 + 2 𝑖)}^{5} = {(2 \sqrt{2})}^{5} 𝑒^{15 𝜋 𝑖 / 4} = 128 \sqrt{2} 𝑒^{- 𝑖 𝜋 / 4} = 128 \sqrt{2} (\frac{1}{\sqrt{2}} - \frac{𝑖}{\sqrt{2}}) = 128 - 128 𝑖,

using ${(2 \sqrt{2})}^{5} = 32 \cdot 4 \sqrt{2}$ and $15 𝜋 / 4 = 4 𝜋 - 𝜋 / 4$ . Also ${(- 1 - \sqrt{3} 𝑖)}^{3} =$ $8 𝑒^{- 2 𝜋 𝑖} =$ $8$ . So the quotient is $(128 - 128 𝑖) / 8 =$ $16 -$ $16 𝑖$ .

B.1.5Roots of unity

Definition B.1.24 (Roots of unity). Let $𝑛 \geq 1$ . An $𝑛$ th root of unity is a complex number $𝜁$ with $𝜁^{𝑛} = 1$ . When $𝑛$ is fixed we write

𝜔 = 𝑒^{2 𝜋 𝑖 / 𝑛} = cos \frac{2 𝜋}{𝑛} + 𝑖 sin \frac{2 𝜋}{𝑛},

and $𝜔_{𝑛}$ when several values of $𝑛$ are in play.

For $𝑛 = 4$ , $𝜔 = 𝑖$ , and the $4$ th roots of unity are $1, 𝑖, - 1, - 𝑖$ . For $𝑛 = 3$ , $𝜔 = - 1 / 2 + (\sqrt{3} / 2) 𝑖$ , and $𝜔^{2} = 𝜔$ . Every $𝑛$ th root of unity is nonzero, and $- 1$ is an $𝑛$ th root of unity exactly when $𝑛$ is even.

Theorem B.1.25 (The $𝑛$ th roots of unity). Let $𝑛 \geq 1$ and $𝜔 = 𝑒^{2 𝜋 𝑖 / 𝑛}$ .

(a)For every integer $𝑚$ , $𝜔^{𝑚} = 𝑒^{2 𝜋 𝑖 𝑚 / 𝑛}$ , and $𝜔^{𝑚} = 1$ if and only if $𝑛 ∣ 𝑚$ . Consequently $𝜔^{𝑚} = 𝜔^{𝑚^{'}}$ if and only if $𝑛 ∣ 𝑚 - 𝑚^{'}$ .
(b)The $𝑛$ th roots of unity are exactly $1, 𝜔, 𝜔^{2}, \dots, 𝜔^{𝑛 - 1}$ , and these $𝑛$ numbers are distinct.
(c)Every $𝑛$ th root of unity $𝜁$ satisfies $| 𝜁 | = 1$ and $𝜁 = 𝜁^{- 1}$ . Products and inverses of $𝑛$ th roots of unity are $𝑛$ th roots of unity.
(d)If $𝑛$ is even, then $𝜔^{𝑛 / 2} = - 1$ .

Proof.standard

Uses. Theorem B.1.20 (De Moivre’s formula), Notation B.1.19 ( $𝑒^{𝑖 𝜃}$ ), Proposition B.1.15 (Existence and uniqueness of polar form), Proposition B.1.8 (Properties of the modulus)

Techniques. Use polar form for products, powers, and roots (Technique B.1.21)

Hint 1. For (a), apply de Moivre’s formula, then ask when

𝑒^{𝑖 𝜑} = 1

Hint 2. For the hard half of (b), let

𝜁^{𝑛} = 1

. First show

| 𝜁 | = 1

, then write

𝜁 = 𝑒^{𝑖 𝜃}

and use the criterion for

𝑒^{𝑖 𝑛 𝜃} = 1

Hint 3. For (c),

{| 𝜁 |}^{𝑛} = | 𝜁^{𝑛} | = 1

, and

𝜁 𝜁 = {| 𝜁 |}^{2}

Sketch. (a)

𝜔^{𝑚} = 𝑒^{2 𝜋 𝑖 𝑚 / 𝑛}

by de Moivre, and

𝑒^{𝑖 𝜑} = 1

exactly when

𝜑 \in 2 𝜋 ℤ

, which here means

𝑚 / 𝑛 \in ℤ

. (b) Each

𝜔^{𝑘}

is a root, since

{(𝜔^{𝑘})}^{𝑛} =

{(𝜔^{𝑛})}^{𝑘} =

1

, and they are distinct by (a). If

𝜁^{𝑛} = 1

, then

| 𝜁 | = 1

, so

𝜁 = 𝑒^{𝑖 𝜃}

, and

𝑒^{𝑖 𝑛 𝜃} = 1

gives

𝑛 𝜃 = 2 𝜋 𝑘

, so

𝜁 = 𝜔^{𝑘}

; reduce

𝑘

modulo

𝑛

. (c)

{| 𝜁 |}^{𝑛} = 1

forces

| 𝜁 | = 1

, so

𝜁 = 𝜁^{- 1}

. (d)

𝜔^{𝑛 / 2} = 𝑒^{𝑖 𝜋} = - 1

(a) By Theorem B.1.20, $𝜔^{𝑚} =$ $𝑒^{𝑖 (2 𝜋 𝑚 / 𝑛)} =$ $𝑒^{2 𝜋 𝑖 𝑚 / 𝑛}$ . By Notation B.1.19 this is $1$ if and only if $2 𝜋 𝑚 / 𝑛 \in 2 𝜋 ℤ$ , that is, $𝑚 / 𝑛 \in ℤ$ , that is, $𝑛 ∣ 𝑚$ . Multiplying by $𝜔^{- 𝑚^{'}}$ , $𝜔^{𝑚} = 𝜔^{𝑚^{'}}$ if and only if $𝜔^{𝑚 - 𝑚^{'}} = 1$ , if and only if $𝑛 ∣ 𝑚 - 𝑚^{'}$ .

(b) For $0 \leq 𝑘 \leq 𝑛 - 1$ , ${(𝜔^{𝑘})}^{𝑛} =$ ${(𝜔^{𝑛})}^{𝑘} =$ $1$ by (a), so each $𝜔^{𝑘}$ is an $𝑛$ th root of unity. If $0 \leq 𝑗 < 𝑘 \leq 𝑛 - 1$ , then $0 < 𝑘 - 𝑗 < 𝑛$ , so $𝑛 ∤ 𝑘 - 𝑗$ , and $𝜔^{𝑗} \neq 𝜔^{𝑘}$ by (a).

Conversely, let $𝜁^{𝑛} = 1$ . Then $𝜁 \neq 0$ , and ${| 𝜁 |}^{𝑛} = | 𝜁^{𝑛} | = 1$ by Proposition B.1.8(d). A positive real number $𝑟$ with $𝑟^{𝑛} = 1$ equals $1$ , since $𝑟 < 1$ gives $𝑟^{𝑛} < 1$ and $𝑟 > 1$ gives $𝑟^{𝑛} > 1$ . So $| 𝜁 | = 1$ , and $𝜁 = 𝑒^{𝑖 𝜃}$ for an argument $𝜃$ of $𝜁$ (Proposition B.1.15). By Theorem B.1.20, $𝑒^{𝑖 𝑛 𝜃} = 𝜁^{𝑛} = 1$ , so $𝑛 𝜃 = 2 𝜋 𝑘$ for some $𝑘 \in ℤ$ , and $𝜁 =$ $𝑒^{2 𝜋 𝑖 𝑘 / 𝑛} =$ $𝜔^{𝑘}$ . Dividing $𝑘$ by $𝑛$ with remainder, $𝑘 = 𝑞 𝑛 + 𝑟$ with $0 \leq 𝑟 \leq 𝑛 - 1$ , and $𝜔^{𝑘} = 𝜔^{𝑟}$ by (a). So $𝜁$ is one of $1, 𝜔, \dots, 𝜔^{𝑛 - 1}$ .

(c) As in (b), $| 𝜁 | = 1$ , so $𝜁 = 𝜁^{- 1}$ by Proposition B.1.8(c). If $𝜁^{𝑛} = 𝜂^{𝑛} = 1$ , then ${(𝜁 𝜂)}^{𝑛} =$ $𝜁^{𝑛} 𝜂^{𝑛} =$ $1$ and ${(𝜁^{- 1})}^{𝑛} =$ ${(𝜁^{𝑛})}^{- 1} =$ $1$ .

(d) By (a), $𝜔^{𝑛 / 2} =$ $𝑒^{2 𝜋 𝑖 (𝑛 / 2) / 𝑛} =$ $𝑒^{𝑖 𝜋} =$ $- 1$ . □

The roots $𝜔^{𝑘} = 𝑒^{2 𝜋 𝑖 𝑘 / 𝑛}$ lie on the unit circle at the angles $2 𝜋 𝑘 / 𝑛$ , equally spaced. So for $𝑛 \geq 3$ they are the vertices of a regular $𝑛$ -gon inscribed in the unit circle, with one vertex at $1$ (Figure B.1.3(a)). Multiplying by $𝜔$ rotates the plane by $2 𝜋 / 𝑛$ (Proposition B.1.18), which carries each vertex $𝜔^{𝑘}$ to the next one, $𝜔^{𝑘 + 1}$ , and $𝜔^{𝑛 - 1}$ back to $𝜔^{𝑛} = 1$ . Parts (a)–(d) are the facts about $𝜔$ behind the discrete Fourier basis (Section 6.3.4).

Figure B.1.3. (a) The

8

th roots of unity

𝜔^{𝑘}

𝜔 = 𝑒^{2 𝜋 𝑖 / 8}

, are the vertices of a regular octagon; multiplication by

𝜔

turns each vertex into the next. (b) The cube roots of

8 𝑖

(Example B.1.27) form an equilateral triangle inscribed in the circle of radius

2

Every nonzero number has $𝑛$ $𝑛$ th roots, by the same method.

Theorem B.1.26 ( $𝑛$ th roots). Let $𝑛 \geq 1$ , and let $𝑤 = 𝑟 𝑒^{𝑖 𝜑}$ with $𝑟 > 0$ and $𝜑 \in ℝ$ . The equation $𝑧^{𝑛} = 𝑤$ has exactly $𝑛$ solutions in $ℂ$ , namely

𝑧_{𝑘} = 𝑟^{1 / 𝑛} 𝑒^{𝑖 (𝜑 + 2 𝜋 𝑘) / 𝑛}, 𝑘 = 0, 1, \dots, 𝑛 - 1,

where $𝑟^{1 / 𝑛}$ is the positive real $𝑛$ th root of $𝑟$ . If $𝑧_{*}$ is any one solution, the solutions are exactly the numbers $𝑧_{*} 𝜁$ with $𝜁$ an $𝑛$ th root of unity. For $𝑛 \geq 3$ they are the vertices of a regular $𝑛$ -gon inscribed in the circle of radius $𝑟^{1 / 𝑛}$ about $0$ .

Proof.standard

Uses. Theorem B.1.20 (De Moivre’s formula), Theorem B.1.25 (The $𝑛$ th roots of unity), Corollary 1.1.14 (Cancellation), Proposition B.1.17 (Multiplication in polar form)

Techniques. Use polar form for products, powers, and roots (Technique B.1.21)

Hint 1. Check with de Moivre’s formula that

𝑧_{0}

is a solution.

Hint 2. If

𝑧_{*}

and

𝑧

are both solutions, what equation does

𝑧 / 𝑧_{*}

satisfy?

Sketch. By de Moivre,

𝑧_{0}^{𝑛} = 𝑟 𝑒^{𝑖 𝜑} = 𝑤

. If

𝑧_{*}

is one solution, then

𝑧^{𝑛} = 𝑤

if and only if

{(𝑧 / 𝑧_{*})}^{𝑛} = 1

, that is,

𝑧 = 𝑧_{*} 𝜁

for an

𝑛

th root of unity

𝜁

, and there are exactly

𝑛

of these by Theorem B.1.25. With

𝑧_{*} = 𝑧_{0}

this is the list

𝑧_{𝑘} = 𝑧_{0} 𝜔^{𝑘}

By Theorem B.1.20, $𝑧_{0}^{𝑛} =$ ${(𝑟^{1 / 𝑛})}^{𝑛} 𝑒^{𝑖 𝜑} =$ $𝑤$ , so solutions exist. Let $𝑧_{*}$ be any solution. Since $𝑤 \neq 0$ , also $𝑧_{*} \neq 0$ . For $𝑧 \in ℂ$ , $𝑧^{𝑛} = 𝑤$ if and only if ${(𝑧 / 𝑧_{*})}^{𝑛} =$ $𝑧^{𝑛} / 𝑤 =$ $1$ , that is, $𝑧 / 𝑧_{*}$ is an $𝑛$ th root of unity. So the solutions are the numbers $𝑧_{*} 𝜁$ with $𝜁^{𝑛} = 1$ . By Theorem B.1.25(b) there are exactly $𝑛$ such $𝜁$ , namely $1, 𝜔, \dots, 𝜔^{𝑛 - 1}$ with $𝜔 = 𝑒^{2 𝜋 𝑖 / 𝑛}$ , and the numbers $𝑧_{*} 𝜔^{𝑘}$ are distinct because $𝑧_{*} \neq 0$ (Corollary 1.1.14). Taking $𝑧_{*} = 𝑧_{0}$ and using Proposition B.1.17,

𝑧_{0} 𝜔^{𝑘} = 𝑟^{1 / 𝑛} 𝑒^{𝑖 𝜑 / 𝑛} 𝑒^{2 𝜋 𝑖 𝑘 / 𝑛} = 𝑟^{1 / 𝑛} 𝑒^{𝑖 (𝜑 + 2 𝜋 𝑘) / 𝑛} = 𝑧_{𝑘} .

All the $𝑧_{𝑘}$ have modulus $𝑟^{1 / 𝑛}$ , and their arguments $(𝜑 + 2 𝜋 𝑘) / 𝑛$ are spaced $2 𝜋 / 𝑛$ apart, so for $𝑛 \geq 3$ they are the vertices of a regular $𝑛$ -gon. □

For $𝑤 = 0$ , the only solution of $𝑧^{𝑛} = 0$ is $𝑧 = 0$ , by Proposition 1.1.17.

Example B.1.27 (Cube roots of

8 𝑖

, fourth roots of

- 4

(a) $8 𝑖 = 8 𝑒^{𝑖 𝜋 / 2}$ , so the cube roots of $8 𝑖$ are $2 𝑒^{𝑖 (𝜋 / 2 + 2 𝜋 𝑘) / 3}$ for $𝑘 = 0, 1, 2$ , with arguments $𝜋 / 6$ , $5 𝜋 / 6$ , and $3 𝜋 / 2$ :
$2 𝑒^{𝑖 𝜋 / 6} = \sqrt{3} + 𝑖, 2 𝑒^{5 𝜋 𝑖 / 6} = - \sqrt{3} + 𝑖, 2 𝑒^{3 𝜋 𝑖 / 2} = - 2 𝑖 .$
For instance, ${(- 2 𝑖)}^{3} = - 8 𝑖^{3} = 8 𝑖$ . They form an equilateral triangle (Figure B.1.3(b)).
(b) $- 4 = 4 𝑒^{𝑖 𝜋}$ , so the fourth roots of $- 4$ are $\sqrt{2} 𝑒^{𝑖 (𝜋 + 2 𝜋 𝑘) / 4}$ for $𝑘 = 0, 1, 2, 3$ . That is, they are $\sqrt{2} 𝑒^{𝑖 𝜋 / 4} = 1 + 𝑖$ times the fourth roots of unity $1, 𝑖, - 1, - 𝑖$ : the numbers $\pm 1 \pm 𝑖$ of Example B.1.5.

The sum of the roots of unity

The $𝑛$ th roots of unity are the vertices of a regular polygon centred at $0$ , so their average, the centre of mass of the vertices, should be $0$ . Rotating by $2 𝜋 / 𝑛$ permutes the vertices, so their sum $𝑆$ satisfies $𝜔 𝑆 = 𝑆$ , and as $𝜔 \neq 1$ , this forces $𝑆 = 0$ . Here are two proofs, together with the product of the roots.

Proposition B.1.28 (Sum and product of the roots of unity). Let $𝑛 \geq 2$ and $𝜔 = 𝑒^{2 𝜋 𝑖 / 𝑛}$ . The $𝑛$ th roots of unity add up to $0$ , and their product is ${(- 1)}^{𝑛 - 1}$ :

\sum_{𝑘 = 0}^{𝑛 - 1} 𝜔^{𝑘} = 0, \prod_{𝑘 = 0}^{𝑛 - 1} 𝜔^{𝑘} = {(- 1)}^{𝑛 - 1} .

Proof (factor and cancel).routine

Uses. Theorem B.1.25 (The $𝑛$ th roots of unity), Proposition 1.1.17 (No zero divisors)

Hint 1. Multiply the sum

𝑆 = 1 + 𝜔 + \dots + 𝜔^{𝑛 - 1}

𝜔 - 1

. The products telescope.

Hint 2. For the product, add the exponents:

0 +

1 +

\dots +

(𝑛 - 1) =

𝑛 (𝑛 - 1) / 2

. Treat odd and even

𝑛

separately, using Theorem B.1.25(a) and (d).

Sketch.

(𝜔 - 1) 𝑆 = 𝜔^{𝑛} - 1 = 0

, and

𝜔 \neq 1

since

𝑛 \geq 2

, so

𝑆 = 0

. The product is

𝜔^{𝑛 (𝑛 - 1) / 2}

. For odd

𝑛

𝑛

divides

𝑛 (𝑛 - 1) / 2

, so the product is

1

; for even

𝑛

, it is

{(𝜔^{𝑛 / 2})}^{𝑛 - 1} =

{(- 1)}^{𝑛 - 1}

Let $𝑆 = 1 + 𝜔 + \dots + 𝜔^{𝑛 - 1}$ . Then

(𝜔 - 1) 𝑆 = (𝜔 + 𝜔^{2} + \dots + 𝜔^{𝑛}) - (1 + 𝜔 + \dots + 𝜔^{𝑛 - 1}) = 𝜔^{𝑛} - 1 = 0 .

Since $𝑛 \geq 2$ , $𝑛 ∤ 1$ , so $𝜔 \neq 1$ by Theorem B.1.25(a), and Proposition 1.1.17 gives $𝑆 = 0$ .

The product is $𝜔^{0 + 1 + \dots + (𝑛 - 1)} =$ $𝜔^{𝑛 (𝑛 - 1) / 2}$ . If $𝑛$ is odd, then $(𝑛 - 1) / 2$ is an integer, so $𝑛$ divides $𝑛 (𝑛 - 1) / 2$ , and the product is $1 = {(- 1)}^{𝑛 - 1}$ by Theorem B.1.25(a). If $𝑛$ is even, then $𝑛 (𝑛 - 1) / 2 =$ $(𝑛 / 2) (𝑛 - 1)$ , and by Theorem B.1.25(d) the product is ${(𝜔^{𝑛 / 2})}^{𝑛 - 1} =$ ${(- 1)}^{𝑛 - 1}$ . □

Proof (compare coefficients).standard

Uses. Theorem B.1.25 (The $𝑛$ th roots of unity), the factor theorem: a polynomial of degree $𝑛$ with leading coefficient $1$ and $𝑛$ distinct roots $𝑐_{1}, \dots, 𝑐_{𝑛}$ equals $(𝑥 - 𝑐_{1}) \dots (𝑥 - 𝑐_{𝑛})$ (Section 7.1)

Techniques. Polynomial identity arguments (Technique 5.5.24)

Hint 1. The

𝑛

th roots of unity are exactly the roots of the polynomial

𝑥^{𝑛} - 1

, and there are

𝑛

of them. Factor

𝑥^{𝑛} - 1

Hint 2. In

(𝑥 - 𝑐_{1}) \dots (𝑥 - 𝑐_{𝑛})

, the coefficient of

𝑥^{𝑛 - 1}

- (𝑐_{1} + \dots + 𝑐_{𝑛})

and the constant term is

{(- 1)}^{𝑛} 𝑐_{1} \dots 𝑐_{𝑛}

Sketch. By Theorem B.1.25,

𝑥^{𝑛} - 1

has the

𝑛

distinct roots

𝜔^{𝑘}

, so

𝑥^{𝑛} -

1 =

\prod_{𝑘} (𝑥 - 𝜔^{𝑘})

. Comparing the coefficients of

𝑥^{𝑛 - 1}

(which is

0

on the left, as

𝑛 \geq 2

) and the constant terms gives

- \sum_{𝑘} 𝜔^{𝑘} = 0

and

{(- 1)}^{𝑛} \prod_{𝑘} 𝜔^{𝑘} = - 1

The polynomial $𝑥^{𝑛} - 1 \in ℂ [𝑥]$ has degree $𝑛$ and leading coefficient $1$ , and by Theorem B.1.25(b) it has the $𝑛$ distinct roots $1, 𝜔, \dots, 𝜔^{𝑛 - 1}$ . By the factor theorem (Section 7.1),

𝑥^{𝑛} - 1 = (𝑥 - 1) (𝑥 - 𝜔) \dots (𝑥 - 𝜔^{𝑛 - 1}) .

Multiplying out the right side, each term picks $𝑥$ or $- 𝜔^{𝑘}$ from each factor. The terms of degree $𝑛 - 1$ pick $- 𝜔^{𝑘}$ from exactly one factor, so the coefficient of $𝑥^{𝑛 - 1}$ is $- (1 + 𝜔 + \dots + 𝜔^{𝑛 - 1})$ . The constant term picks $- 𝜔^{𝑘}$ from every factor, so it is ${(- 1)}^{𝑛} 𝜔^{0} 𝜔^{1} \dots 𝜔^{𝑛 - 1}$ . On the left, the coefficient of $𝑥^{𝑛 - 1}$ is $0$ , since $𝑛 - 1 \geq 1$ , and the constant term is $- 1$ . Comparing,

\sum_{𝑘 = 0}^{𝑛 - 1} 𝜔^{𝑘} = 0, \prod_{𝑘 = 0}^{𝑛 - 1} 𝜔^{𝑘} = {(- 1)}^{𝑛 + 1} = {(- 1)}^{𝑛 - 1} .

□

The first proof needs nothing about polynomials. The second reads both identities off one factorization, the way Vieta’s formulas relate the coefficients of a polynomial to its roots, and the other coefficients of $𝑥^{𝑛} - 1$ give more identities of the same kind (Exercise B.1.45 uses the factorization again).

The first proof applies to any $𝑛$ th root of unity $𝜁 \neq 1$ : since $(𝜁 - 1) (1 + 𝜁 + \dots + 𝜁^{𝑛 - 1}) =$ $𝜁^{𝑛} -$ $1 =$ $0$ , we get $1 + 𝜁 + \dots + 𝜁^{𝑛 - 1} = 0$ . With $𝜁 = 𝜔^{𝑚}$ this is Lemma 6.3.39, the computation behind the orthogonality of the discrete Fourier basis.

B.1.6Square roots and quadratic equations

$ℝ$ was enlarged to $ℂ$ to give $- 1$ a square root. Every complex number then has a square root without any further enlargement, and so every quadratic equation over $ℂ$ can be solved.

Proposition B.1.29 (Square roots in

ℂ

). Every

𝑤 \in ℂ

has a square root in

ℂ

. If

𝑤 \neq 0

, it has exactly two square roots,

𝑧_{*}

and

- 𝑧_{*}

, where

𝑧_{*}

is any one of them. The only square root of

0

0

Proof (by polar form).routine

Uses. Theorem B.1.26 ( $𝑛$ th roots), Theorem B.1.25 (The $𝑛$ th roots of unity), Proposition 1.1.17 (No zero divisors)

Techniques. Use polar form for products, powers, and roots (Technique B.1.21)

Hint 1. Apply Theorem B.1.26 with

𝑛 = 2

Hint 2. The square roots of unity are

1

and

𝑒^{𝑖 𝜋} = - 1

Sketch. For

𝑤 \neq 0

, Theorem B.1.26 with

𝑛 = 2

gives exactly the two solutions

𝑧_{*} 𝜁

with

𝜁 \in {1, - 1}

. For

𝑤 = 0

𝑧^{2} = 0

forces

𝑧 = 0

𝑤 = 0

, then

𝑧^{2} = 0

forces

𝑧 = 0

(Proposition 1.1.17). If

𝑤 \neq 0

, Theorem B.1.26 with

𝑛 = 2

gives exactly two solutions of

𝑧^{2} = 𝑤

, namely

𝑧_{*} 𝜁

for the two square roots of unity

𝜁

. By Theorem B.1.25 with

𝑛 = 2

, these are

1

and

𝜔^{2 / 2} = - 1

, so the solutions are

𝑧_{*}

and

- 𝑧_{*}

. Explicitly, if

𝑤 = 𝑟 𝑒^{𝑖 𝜑}

, then

𝑧_{*} = \sqrt{𝑟} 𝑒^{𝑖 𝜑 / 2}

is one of them. □

Proof (by solving for the real and imaginary parts).standard

Uses. Proposition B.1.8 (Properties of the modulus), Proposition 1.1.17 (No zero divisors), Definition 1.1.29 (Complex numbers)

Hint 1. Write

𝑤 = 𝑎 + 𝑏 𝑖

and look for

𝑧 = 𝑥 + 𝑦 𝑖

. Then

𝑧^{2} = 𝑤

says

𝑥^{2} - 𝑦^{2} = 𝑎

and

2 𝑥 𝑦 = 𝑏

Hint 2. Taking moduli,

𝑥^{2} + 𝑦^{2} = | 𝑤 |

. Combine this with

𝑥^{2} - 𝑦^{2} = 𝑎

to find

𝑥^{2}

and

𝑦^{2}

, then fix the signs with

2 𝑥 𝑦 = 𝑏

Hint 3. For the count, factor

𝑧^{2} - 𝑧_{*}^{2}

Sketch. A solution must satisfy

𝑥^{2} - 𝑦^{2} = 𝑎

and

𝑥^{2} + 𝑦^{2} = | 𝑤 |

, so

𝑥^{2} = (| 𝑤 | + 𝑎) / 2

and

𝑦^{2} = (| 𝑤 | - 𝑎) / 2

, both nonnegative because

| 𝑎 | \leq | 𝑤 |

. Taking the nonnegative root for

𝑥

and giving

𝑦

the sign of

𝑏

(either sign if

𝑏 = 0

), one checks that

𝑥^{2} - 𝑦^{2} = 𝑎

and

2 𝑥 𝑦 = 𝑏

. Finally,

𝑧^{2} = 𝑧_{*}^{2}

if and only if

(𝑧 - 𝑧_{*}) (𝑧 + 𝑧_{*}) =

0

Let $𝑤 = 𝑎 + 𝑏 𝑖$ . We first find a candidate. If $𝑧 = 𝑥 + 𝑦 𝑖$ satisfies $𝑧^{2} = 𝑤$ , then $𝑥^{2} - 𝑦^{2} = 𝑎$ and $2 𝑥 𝑦 = 𝑏$ , and taking moduli, $𝑥^{2} +$ $𝑦^{2} =$ ${| 𝑧 |}^{2} =$ $| 𝑧^{2} | =$ $| 𝑤 |$ (Proposition B.1.8(d)). Hence

𝑥^{2} = \frac{| 𝑤 | + 𝑎}{2}, 𝑦^{2} = \frac{| 𝑤 | - 𝑎}{2} .

Both right sides are nonnegative, since $| 𝑎 | \leq | 𝑤 |$ (Proposition B.1.8(e)). So let $𝜎 = 1$ if $𝑏 \geq 0$ and $𝜎 = - 1$ if $𝑏 < 0$ , and put

𝑧_{*} = \sqrt{\frac{| 𝑤 | + 𝑎}{2}} + 𝜎 \sqrt{\frac{| 𝑤 | - 𝑎}{2}} 𝑖 .

(B.1.3)

We check that $𝑧_{*}^{2} = 𝑤$ . The real part of $𝑧_{*}^{2}$ is $(| 𝑤 | + 𝑎) / 2 -$ $(| 𝑤 | - 𝑎) / 2 =$ $𝑎$ , and its imaginary part is

2 𝜎 \sqrt{\frac{| 𝑤 | + 𝑎}{2}} \sqrt{\frac{| 𝑤 | - 𝑎}{2}} = 𝜎 \sqrt{{| 𝑤 |}^{2} - 𝑎^{2}} = 𝜎 \sqrt{𝑏^{2}} = 𝜎 | 𝑏 | = 𝑏 .

So every $𝑤$ has a square root. Now if $𝑧_{*}^{2} = 𝑤$ , then $𝑧^{2} = 𝑤$ if and only if $(𝑧 - 𝑧_{*}) (𝑧 + 𝑧_{*}) =$ $𝑧^{2} -$ $𝑧_{*}^{2} =$ $0$ , which by Proposition 1.1.17 holds if and only if $𝑧 = 𝑧_{*}$ or $𝑧 = - 𝑧_{*}$ . These differ unless $2 𝑧_{*} = 0$ , that is, unless $𝑧_{*} = 0$ , which happens exactly when $𝑤 = 0$ . □

The second proof needs no trigonometry: (B.1.3) computes a square root from two real square roots. For $𝑤 = 3 + 4 𝑖$ we have $| 𝑤 | = 5$ , so $𝑧_{*} = \sqrt{4} + \sqrt{1} 𝑖 = 2 + 𝑖$ , and indeed ${(2 + 𝑖)}^{2} = 3 + 4 𝑖$ . For $𝑤 = 𝑖$ it gives $𝑧_{*} = (1 + 𝑖) / \sqrt{2}$ , as in Exercise 1.1.44(b).

Theorem B.1.30 (Quadratic equations over $ℂ$ ). Let $𝑎, 𝑏, 𝑐 \in ℂ$ with $𝑎 \neq 0$ , let $Δ = 𝑏^{2} - 4 𝑎 𝑐$ , and let $𝑠$ be a square root of $Δ$ (Proposition B.1.29). Then the solutions of $𝑎 𝑧^{2} + 𝑏 𝑧 + 𝑐 = 0$ are

𝑧_{1} = \frac{- 𝑏 + 𝑠}{2 𝑎}, 𝑧_{2} = \frac{- 𝑏 - 𝑠}{2 𝑎},

which are distinct if $Δ \neq 0$ and equal if $Δ = 0$ . Moreover, $𝑎 𝑧^{2} +$ $𝑏 𝑧 +$ $𝑐 =$ $𝑎 (𝑧 - 𝑧_{1}) (𝑧 - 𝑧_{2})$ for every $𝑧 \in ℂ$ .

Proof.routine

Uses. Proposition B.1.29 (Square roots in $ℂ$ ), Exercise 1.1.50

Hint 1. Complete the square: multiply the equation by

4 𝑎

Hint 2.

4 𝑎 (𝑎 𝑧^{2} + 𝑏 𝑧 + 𝑐) =

{(2 𝑎 𝑧 + 𝑏)}^{2} -

Δ

. Now use that the square roots of

Δ

are

\pm 𝑠

Hint 3. For the factorization, compute

𝑧_{1} + 𝑧_{2}

and

𝑧_{1} 𝑧_{2}

Sketch. Since

4 𝑎 (𝑎 𝑧^{2} + 𝑏 𝑧 + 𝑐) =

{(2 𝑎 𝑧 + 𝑏)}^{2} -

Δ

and

4 𝑎 \neq 0

𝑧

is a solution if and only if

2 𝑎 𝑧 + 𝑏

is a square root of

Δ

, that is,

2 𝑎 𝑧 + 𝑏 = \pm 𝑠

. The roots coincide exactly when

𝑠 = 0

. Finally

𝑧_{1} + 𝑧_{2} = - 𝑏 / 𝑎

and

𝑧_{1} 𝑧_{2} =

(𝑏^{2} - 𝑠^{2}) / (4 𝑎^{2}) =

𝑐 / 𝑎

Expanding, ${(2 𝑎 𝑧 + 𝑏)}^{2} -$ $Δ =$ $4 𝑎^{2} 𝑧^{2} +$ $4 𝑎 𝑏 𝑧 +$ $4 𝑎 𝑐 =$ $4 𝑎 (𝑎 𝑧^{2} + 𝑏 𝑧 + 𝑐)$ . Since $4 𝑎 \neq 0$ , $𝑧$ is a solution if and only if ${(2 𝑎 𝑧 + 𝑏)}^{2} = Δ$ , that is, $2 𝑎 𝑧 + 𝑏$ is a square root of $Δ$ . By Proposition B.1.29 the square roots of $Δ$ are $𝑠$ and $- 𝑠$ , so $𝑧$ is a solution if and only if $2 𝑎 𝑧 + 𝑏 = \pm 𝑠$ , that is, $𝑧 = 𝑧_{1}$ or $𝑧 = 𝑧_{2}$ . These coincide if and only if $𝑠 = - 𝑠$ , that is, $𝑠 = 0$ , that is, $Δ = 0$ . Finally $𝑧_{1} + 𝑧_{2} = - 𝑏 / 𝑎$ and

𝑧_{1} 𝑧_{2} = \frac{{(- 𝑏)}^{2} - 𝑠^{2}}{4 𝑎^{2}} = \frac{4 𝑎 𝑐}{4 𝑎^{2}} = \frac{𝑐}{𝑎},

so $𝑎 (𝑧 - 𝑧_{1}) (𝑧 - 𝑧_{2}) =$ $𝑎 𝑧^{2} -$ $𝑎 (𝑧_{1} + 𝑧_{2}) 𝑧 +$ $𝑎 𝑧_{1} 𝑧_{2} =$ $𝑎 𝑧^{2} +$ $𝑏 𝑧 +$ $𝑐$ . □

This is Exercise 1.1.50(a) for the field $ℂ$ , where the hypothesis that $𝑏^{2} - 4 𝑎 𝑐$ is a square always holds.

Example B.1.31 (A quadratic with complex coefficients). To solve $𝑧^{2} - 3 𝑧 + (3 - 𝑖) = 0$ , compute $Δ =$ $9 -$ $4 (3 - 𝑖) =$ $- 3 +$ $4 𝑖$ . Here $| Δ | = 5$ , and (B.1.3) gives the square root $𝑠 = \sqrt{1} + \sqrt{4} 𝑖 = 1 + 2 𝑖$ . So the solutions are

𝑧 = \frac{3 \pm (1 + 2 𝑖)}{2}, that is, 𝑧 = 2 + 𝑖 and 𝑧 = 1 - 𝑖 .

Check: their sum is $3$ and their product is $(2 + 𝑖) (1 - 𝑖) = 3 - 𝑖$ . The roots are not conjugate to each other, which is allowed because the coefficients are not real (compare Proposition B.1.4).

Remark B.1.32 (The Fundamental Theorem of Algebra). Theorem B.1.26 and Theorem B.1.30 solve two kinds of polynomial equations over

ℂ

. The Fundamental Theorem of Algebra (Theorem 7.1.32) says that every nonconstant polynomial with complex coefficients has a root in

ℂ

. Over

ℝ

this fails already for

𝑥^{2} + 1

, and

ℂ

was built by adjoining a root of that single polynomial; the theorem says that no polynomial ever forces a further enlargement. Together with the factor theorem, it factors every polynomial in

ℂ [𝑥]

into linear factors, and together with Proposition B.1.4, it factors every polynomial in

ℝ [𝑥]

into real factors of degree at most

2

(Section 7.1). For degree

5

and higher there is no formula in radicals like Theorem B.1.30, and the proofs in Section C.1 show that a root exists without producing it. In linear algebra, the theorem is what guarantees that every operator on a nonzero finite-dimensional complex vector space has an eigenvalue (Theorem 7.3.27).

Exercises

Exercise B.1.33.routine Write each of the following in the form

𝑎 + 𝑏 𝑖

: (a)

{(1 - 𝑖)}^{10}

; (b)

{(1 + \sqrt{3} 𝑖)}^{- 4}

; (c)

{(cos (𝜋 / 9) + 𝑖 sin (𝜋 / 9))}^{12}

; (d)

{((1 + 𝑖) / (1 - 𝑖))}^{2027}

Solution.

Uses. Proposition B.1.15 (Existence and uniqueness of polar form), Theorem B.1.20 (De Moivre’s formula), Exercise 1.1.36

Techniques. Use polar form for products, powers, and roots (Technique B.1.21)

Hint 1. Write each base in polar form and apply Theorem B.1.20. Then reduce the argument modulo

2 𝜋

Hint 2.

1 - 𝑖 = \sqrt{2} 𝑒^{- 𝑖 𝜋 / 4}

and

1 + \sqrt{3} 𝑖 = 2 𝑒^{𝑖 𝜋 / 3}

. For (d), first simplify the base; it is a fourth root of unity.

Sketch. (a)

32 𝑒^{- 5 𝜋 𝑖 / 2} =

32 𝑒^{- 𝑖 𝜋 / 2} =

- 32 𝑖

. (b)

2^{- 4} 𝑒^{- 4 𝜋 𝑖 / 3} =

(1 / 16) 𝑒^{2 𝜋 𝑖 / 3} =

- 1 / 32 +

(\sqrt{3} / 32) 𝑖

. (c)

𝑒^{4 𝜋 𝑖 / 3} =

- 1 / 2 -

(\sqrt{3} / 2) 𝑖

. (d) The base is

𝑖

, and

𝑖^{2027} = 𝑖^{3} = - 𝑖

(a) $1 - 𝑖 = \sqrt{2} 𝑒^{- 𝑖 𝜋 / 4}$ , so ${(1 - 𝑖)}^{10} =$ $2^{5} 𝑒^{- 10 𝜋 𝑖 / 4} =$ $32 𝑒^{- 5 𝜋 𝑖 / 2}$ . Since $- 5 𝜋 / 2 = - 𝜋 / 2 - 2 𝜋$ , this is $32 𝑒^{- 𝑖 𝜋 / 2} = - 32 𝑖$ .

(b) $1 + \sqrt{3} 𝑖 = 2 𝑒^{𝑖 𝜋 / 3}$ , so ${(1 + \sqrt{3} 𝑖)}^{- 4} =$ $2^{- 4} 𝑒^{- 4 𝜋 𝑖 / 3} =$ $(1 / 16) 𝑒^{2 𝜋 𝑖 / 3} =$ $(1 / 16) (- 1 / 2 + (\sqrt{3} / 2) 𝑖) =$ $- 1 / 32 +$ $(\sqrt{3} / 32) 𝑖$ .

(c) By Theorem B.1.20, the power is $𝑒^{12 𝜋 𝑖 / 9} =$ $𝑒^{4 𝜋 𝑖 / 3} =$ $- 1 / 2 -$ $(\sqrt{3} / 2) 𝑖$ .

(d) By Exercise 1.1.36(c), $(1 + 𝑖) / (1 - 𝑖) = 𝑖$ . Since $𝑖^{4} = 1$ and $2027 = 4 \cdot 506 + 3$ , the power is $𝑖^{3} = - 𝑖$ .

Exercise B.1.34.routine Find all

𝑧 \in ℂ

with

𝑧^{6} = - 64

, in the form

𝑎 + 𝑏 𝑖

, and draw them. Which of them satisfy

𝑧^{2} = - 4

Solution.

Uses. Theorem B.1.26 ( $𝑛$ th roots), Proposition B.1.4 (Roots of real polynomials come in conjugate pairs)

Techniques. Use polar form for products, powers, and roots (Technique B.1.21)

Hint 1.

- 64 = 64 𝑒^{𝑖 𝜋}

. Apply Theorem B.1.26 with

𝑛 = 6

Hint 2. The arguments are

𝜋 / 6 + 𝑘 𝜋 / 3

for

𝑘 = 0, \dots, 5

, and the modulus is

2

Sketch. The solutions are

2 𝑒^{𝑖 (𝜋 + 2 𝜋 𝑘) / 6}

\pm \sqrt{3} \pm 𝑖

and

\pm 2 𝑖

, the vertices of a regular hexagon of radius

2

. Of these,

\pm 2 𝑖

satisfy

𝑧^{2} = - 4

Since $- 64 = 64 𝑒^{𝑖 𝜋}$ and $64^{1 / 6} = 2$ , Theorem B.1.26 gives the six solutions $2 𝑒^{𝑖 (𝜋 + 2 𝜋 𝑘) / 6}$ , $𝑘 = 0, \dots, 5$ , with arguments $𝜋 / 6,$ $𝜋 / 2,$ $5 𝜋 / 6,$ $7 𝜋 / 6,$ $3 𝜋 / 2,$ $11 𝜋 / 6$ :

\sqrt{3} + 𝑖, 2 𝑖, - \sqrt{3} + 𝑖, - \sqrt{3} - 𝑖, - 2 𝑖, \sqrt{3} - 𝑖 .

They are the vertices of a regular hexagon inscribed in the circle of radius $2$ , with vertices on the imaginary axis. As $𝑧^{6} + 64$ has real coefficients, they come in conjugate pairs (Proposition B.1.4). The solutions of $𝑧^{2} = - 4$ are $\pm 2 𝑖$ , and both are on the list, as they must be: if $𝑧^{2} = - 4$ , then $𝑧^{6} = {(𝑧^{2})}^{3} = - 64$ .

Exercise B.1.35.routine Let $𝑧, 𝑤 \in ℂ$ .

(a)Show that ${| 𝑧 + 𝑤 |}^{2} +$ ${| 𝑧 - 𝑤 |}^{2} =$ $2 {| 𝑧 |}^{2} +$ $2 {| 𝑤 |}^{2}$ , and interpret this in the parallelogram with vertices $0, 𝑧, 𝑤, 𝑧 + 𝑤$ .
(b)Show that ${| 𝑧 + 𝑤 |}^{2} -$ ${| 𝑧 - 𝑤 |}^{2} =$ $4 Re (𝑧 𝑤)$ . Deduce that the two diagonals of the parallelogram have the same length if and only if $Re (𝑧 𝑤) = 0$ , and interpret this condition geometrically.

Solution.

Uses. Proposition B.1.8 (Properties of the modulus), Proposition B.1.2 (Conjugation respects the field operations)

Techniques. Square the modulus (Technique B.1.9)

Hint 1. Apply Proposition B.1.8(f) to

𝑧 + 𝑤

and to

𝑧 + (- 𝑤)

Hint 2.

Re (𝑧 𝑤) =

𝑎 𝑐 +

𝑏 𝑑

for

𝑧 = 𝑎 + 𝑏 𝑖

and

𝑤 = 𝑐 + 𝑑 𝑖

Sketch.

{| 𝑧 \pm 𝑤 |}^{2} =

{| 𝑧 |}^{2} \pm 2 Re (𝑧 𝑤) +

{| 𝑤 |}^{2}

. Adding gives (a): the squares of the diagonals add up to the squares of the four sides. Subtracting gives (b), and

Re (𝑧 𝑤) =

𝑎 𝑐 +

𝑏 𝑑

is the dot product of

(𝑎, 𝑏)

and

(𝑐, 𝑑)

, so the diagonals are equal exactly when the parallelogram is a rectangle.

By Proposition B.1.8(f), ${| 𝑧 + 𝑤 |}^{2} =$ ${| 𝑧 |}^{2} +$ $2 Re (𝑧 𝑤) +$ ${| 𝑤 |}^{2}$ . Applied to $𝑧$ and $- 𝑤$ , with $- 𝑤 = - 𝑤$ and $| - 𝑤 | = | 𝑤 |$ , it gives ${| 𝑧 - 𝑤 |}^{2} =$ ${| 𝑧 |}^{2} -$ $2 Re (𝑧 𝑤) +$ ${| 𝑤 |}^{2}$ .

(a) Adding the two identities gives ${| 𝑧 + 𝑤 |}^{2} +$ ${| 𝑧 - 𝑤 |}^{2} =$ $2 {| 𝑧 |}^{2} +$ $2 {| 𝑤 |}^{2}$ . The diagonals of the parallelogram are the segments from $0$ to $𝑧 + 𝑤$ and from $𝑤$ to $𝑧$ , of lengths $| 𝑧 + 𝑤 |$ and $| 𝑧 - 𝑤 |$ , and its four sides have lengths $| 𝑧 |,$ $| 𝑤 |,$ $| 𝑧 |,$ $| 𝑤 |$ . So the sum of the squares of the diagonals equals the sum of the squares of the sides. This is the parallelogram law, the case $𝑉 = ℂ$ of Proposition 6.1.30.

(b) Subtracting gives ${| 𝑧 + 𝑤 |}^{2} -$ ${| 𝑧 - 𝑤 |}^{2} =$ $4 Re (𝑧 𝑤)$ . Both lengths are nonnegative, so $| 𝑧 + 𝑤 | = | 𝑧 - 𝑤 |$ if and only if $Re (𝑧 𝑤) = 0$ . For $𝑧 = 𝑎 + 𝑏 𝑖$ and $𝑤 = 𝑐 + 𝑑 𝑖$ , $𝑧 𝑤 =$ $(𝑎 𝑐 + 𝑏 𝑑) +$ $(𝑏 𝑐 - 𝑎 𝑑) 𝑖$ , so $Re (𝑧 𝑤) =$ $𝑎 𝑐 +$ $𝑏 𝑑$ is the dot product of $(𝑎, 𝑏)$ and $(𝑐, 𝑑)$ . So the diagonals are equally long exactly when the sides $𝑧$ and $𝑤$ are perpendicular (or one of them is $0$ ), that is, when the parallelogram is a rectangle.

Exercise B.1.36.standard Let $𝑧, 𝑤 \in ℂ$ .

(a)Show that ${| 1 - 𝑧 𝑤 |}^{2} -$ ${| 𝑧 - 𝑤 |}^{2} =$ $(1 - {| 𝑧 |}^{2}) (1 - {| 𝑤 |}^{2})$ .
(b)Deduce that if $| 𝑧 | < 1$ and $| 𝑤 | < 1$ , then $1 - 𝑧 𝑤 \neq 0$ and
$| \frac{𝑧 - 𝑤}{1 - 𝑧 𝑤} | < 1 .$
(c)Show that if $| 𝑤 | = 1$ and $𝑧 𝑤 \neq 1$ , then $| (𝑧 - 𝑤) / (1 - 𝑧 𝑤) | =$ $1$ .

Solution.

Uses. Proposition B.1.8 (Properties of the modulus), Proposition B.1.2 (Conjugation respects the field operations)

Techniques. Square the modulus (Technique B.1.9)

Hint 1. Expand both squared moduli with Proposition B.1.8(f), or as

𝑢 𝑢

Hint 2. The cross terms are

- 2 Re (𝑧 𝑤)

and

- 2 Re (𝑧 𝑤)

. These are equal, because

𝑧 𝑤

is the conjugate of

𝑧 𝑤

Hint 3. For (b), the right side of (a) is positive. For (c), it is zero.

Sketch.

{| 1 - 𝑧 𝑤 |}^{2} =

1 -

2 Re (𝑧 𝑤) +

{| 𝑧 |}^{2} {| 𝑤 |}^{2}

and

{| 𝑧 - 𝑤 |}^{2} =

{| 𝑧 |}^{2} -

2 Re (𝑧 𝑤) +

{| 𝑤 |}^{2}

; the real parts agree, and the difference is

1 -

{| 𝑧 |}^{2} -

{| 𝑤 |}^{2} +

{| 𝑧 |}^{2} {| 𝑤 |}^{2}

. In (b),

| 𝑧 𝑤 | < 1

, so the denominator is nonzero, and the identity gives

{| 𝑧 - 𝑤 |}^{2} <

{| 1 - 𝑧 𝑤 |}^{2}

. In (c) the right side vanishes.

(a) By Proposition B.1.8(f) (applied to $1$ and $- 𝑧 𝑤$ , and to $𝑧$ and $- 𝑤$ ) and (d),

\begin{aligned} {| 1 - 𝑧 𝑤 |}^{2} & = 1 - 2 Re (𝑧 𝑤) + {| 𝑧 |}^{2} {| 𝑤 |}^{2}, \\ {| 𝑧 - 𝑤 |}^{2} & = {| 𝑧 |}^{2} - 2 Re (𝑧 𝑤) + {| 𝑤 |}^{2} . \end{aligned}

Since $𝑧 𝑤 = 𝑧 𝑤$ (Proposition B.1.2), the two numbers have the same real part. Subtracting,

{| 1 - 𝑧 𝑤 |}^{2} - {| 𝑧 - 𝑤 |}^{2} = 1 - {| 𝑧 |}^{2} - {| 𝑤 |}^{2} + {| 𝑧 |}^{2} {| 𝑤 |}^{2} = (1 - {| 𝑧 |}^{2}) (1 - {| 𝑤 |}^{2}) .

(b) $| 𝑧 𝑤 | =$ $| 𝑧 | | 𝑤 | <$ $1$ , so $𝑧 𝑤 \neq 1$ and $1 - 𝑧 𝑤 \neq 0$ . Both factors on the right of (a) are positive, so ${| 𝑧 - 𝑤 |}^{2} <$ ${| 1 - 𝑧 𝑤 |}^{2}$ . Taking square roots and dividing by $| 1 - 𝑧 𝑤 | > 0$ gives the claim (Proposition B.1.8(d)).

(c) Now $1 - {| 𝑤 |}^{2} = 0$ , so (a) gives $| 1 - 𝑧 𝑤 | = | 𝑧 - 𝑤 |$ , and the denominator is nonzero by assumption. Hence the quotient has modulus $1$ .

So for fixed $𝑧$ with $| 𝑧 | < 1$ , the map $𝑤 \mapsto$ $(𝑧 - 𝑤) / (1 - 𝑧 𝑤)$ sends the open unit disk into itself, by (b), and the unit circle into itself, by (c); the condition $𝑧 𝑤 \neq 1$ holds automatically, since $| 𝑧 𝑤 | < 1$ .

Exercise B.1.37.standard Describe each of the following sets geometrically.

(a) ${𝑧 \in ℂ | | 𝑧 - 1 | = | 𝑧 + 𝑖 |}$ .
(b) ${𝑧 \in ℂ | | 𝑧 - 1 | = 2 | 𝑧 + 1 |}$ .
(c) ${𝑧 \in ℂ | 𝑧 \neq - 1,$ $Re ((𝑧 - 1) / (𝑧 + 1)) =$ $0}$ .

Solution.

Uses. Proposition B.1.8 (Properties of the modulus), Proposition B.1.2 (Conjugation respects the field operations)

Techniques. Square the modulus (Technique B.1.9)

Hint 1. For (a) and (b), square both sides and write

𝑧 = 𝑥 + 𝑦 𝑖

Hint 2. For (c), multiply the numerator and the denominator by

𝑧 + 1

. The denominator becomes a positive real number, and the numerator is

(𝑧 - 1) (𝑧 + 1)

Sketch. (a) Squaring,

- 2 𝑥 + 1 = 2 𝑦 + 1

: the line

𝑦 = - 𝑥

, the perpendicular bisector of

1

and

- 𝑖

. (b) Squaring gives

3 𝑥^{2} +

10 𝑥 +

3 +

3 𝑦^{2} =

0

, the circle with centre

- 5 / 3

and radius

4 / 3

. (c)

(𝑧 - 1) (𝑧 + 1) =

{| 𝑧 |}^{2} -

1 +

(𝑧 - 𝑧)

has real part

{| 𝑧 |}^{2} - 1

, so the set is the unit circle without

- 1

Write $𝑧 = 𝑥 + 𝑦 𝑖$ . Moduli are nonnegative, so each equation between moduli is equivalent to the equation between their squares.

(a) ${| 𝑧 - 1 |}^{2} =$ ${(𝑥 - 1)}^{2} +$ $𝑦^{2}$ and ${| 𝑧 + 𝑖 |}^{2} =$ $𝑥^{2} +$ ${(𝑦 + 1)}^{2}$ . They are equal if and only if $- 2 𝑥 + 1 = 2 𝑦 + 1$ , that is, $𝑦 = - 𝑥$ . The set is the line through $0$ with slope $- 1$ : the perpendicular bisector of the segment from $1$ to $- 𝑖$ , as it should be, since it consists of the points equidistant from $1$ and $- 𝑖$ .

(b) The condition is ${(𝑥 - 1)}^{2} +$ $𝑦^{2} =$ $4 ({(𝑥 + 1)}^{2} + 𝑦^{2})$ , that is, $3 𝑥^{2} +$ $10 𝑥 +$ $3 +$ $3 𝑦^{2} =$ $0$ . Dividing by $3$ and completing the square, ${(𝑥 + 5 / 3)}^{2} +$ $𝑦^{2} =$ $25 / 9 -$ $1 =$ $16 / 9$ . The set is the circle with centre $- 5 / 3$ and radius $4 / 3$ .

(c) For $𝑧 \neq - 1$ ,

\frac{𝑧 - 1}{𝑧 + 1} = \frac{(𝑧 - 1) (𝑧 + 1)}{{| 𝑧 + 1 |}^{2}} = \frac{(𝑧 - 1) (𝑧 + 1)}{{| 𝑧 + 1 |}^{2}}

by Proposition B.1.8(c), and ${| 𝑧 + 1 |}^{2} > 0$ is real. The numerator is $𝑧 𝑧 +$ $𝑧 -$ $𝑧 -$ $1 =$ $({| 𝑧 |}^{2} - 1) +$ $(𝑧 - 𝑧)$ , where $𝑧 - 𝑧 = 2 𝑦 𝑖$ is purely imaginary. So the real part of the quotient is $({| 𝑧 |}^{2} - 1) / {| 𝑧 + 1 |}^{2}$ , which is $0$ if and only if $| 𝑧 | = 1$ . The set is the unit circle with $- 1$ removed. Geometrically, $(𝑧 - 1) / (𝑧 + 1)$ is purely imaginary exactly when the segments from $𝑧$ to $1$ and to $- 1$ are perpendicular (or $𝑧 = 1$ ), and the set of such $𝑧$ is the circle with diameter from $- 1$ to $1$ (Thales).

Exercise B.1.38.routine Prove or disprove each statement, for all $𝑧, 𝑤 \in ℂ$ .

(a) $Re (𝑧 𝑤) = Re 𝑧 Re 𝑤$ .
(b) $| 𝑧^{2} | = {| 𝑧 |}^{2}$ .
(c)If $𝑧^{2}$ is real, then $𝑧$ is real.
(d)If $| 𝑧 | = | 𝑤 |$ , then $𝑧 = 𝑤$ or $𝑧 = - 𝑤$ .
(e)If $| 𝑧 + 𝑤 | = | 𝑧 - 𝑤 |$ , then $Re (𝑧 𝑤) = 0$ .
(f)If $𝑧 \neq 0$ , then $1 / 𝑧 = 1 / 𝑧$ .

Solution.

Uses. Proposition B.1.8 (Properties of the modulus), Proposition B.1.2 (Conjugation respects the field operations)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Test each claim on

𝑧, 𝑤 \in {1, 𝑖, 1 + 𝑖}

before trying to prove it.

Hint 2. (b), (e), and (f) are true; for (e), expand

{| 𝑧 \pm 𝑤 |}^{2}

with Proposition B.1.8(f).

Sketch. (a) False:

𝑧 = 𝑤 = 𝑖

. (b) True:

| 𝑧 𝑧 | = | 𝑧 | | 𝑧 |

. (c) False:

𝑧 = 𝑖

. (d) False:

𝑧 = 1

𝑤 = 𝑖

. (e) True:

{| 𝑧 + 𝑤 |}^{2} -

{| 𝑧 - 𝑤 |}^{2} =

4 Re (𝑧 𝑤)

. (f) True: Proposition B.1.2(e).

(a) False. For $𝑧 = 𝑤 = 𝑖$ , $Re (𝑧 𝑤) =$ $Re (- 1) =$ $- 1$ , but $Re 𝑧 Re 𝑤 = 0$ . The correct formula is $Re (𝑧 𝑤) =$ $Re 𝑧 Re 𝑤 -$ $Im 𝑧 Im 𝑤$ .

(b) True, by Proposition B.1.8(d) with $𝑤 = 𝑧$ .

(c) False: $𝑖^{2} = - 1$ is real, but $𝑖$ is not.

(d) False: $| 1 | = | 𝑖 | = 1$ , but $1 \neq \pm 𝑖$ . Numbers of equal modulus differ by a factor of modulus $1$ , of which $\pm 1$ are only two.

(e) True. By Proposition B.1.8(f), ${| 𝑧 \pm 𝑤 |}^{2} =$ ${| 𝑧 |}^{2} \pm 2 Re (𝑧 𝑤) +$ ${| 𝑤 |}^{2}$ , so the hypothesis gives $4 Re (𝑧 𝑤) = 0$ .

(f) True, by Proposition B.1.2(e).

Exercise B.1.39.routine Find the error in the following argument.

Claim. $1 = - 1$ .

Proof. By de Moivre’s formula with $𝑛 = 1 / 2$ , ${(cos 𝜃 + 𝑖 sin 𝜃)}^{1 / 2} =$ $cos (𝜃 / 2) +$ $𝑖 sin (𝜃 / 2)$ for every $𝜃 \in ℝ$ . Now $1 =$ $cos 0 +$ $𝑖 sin 0 =$ $cos 2 𝜋 +$ $𝑖 sin 2 𝜋$ . Taking $𝜃 = 0$ gives $1^{1 / 2} =$ $cos 0 +$ $𝑖 sin 0 =$ $1$ , and taking $𝜃 = 2 𝜋$ gives $1^{1 / 2} =$ $cos 𝜋 +$ $𝑖 sin 𝜋 =$ $- 1$ . So $1 = - 1$ .

Solution.

Uses. Theorem B.1.20 (De Moivre’s formula), Proposition B.1.29 (Square roots in $ℂ$ ), Proposition B.1.15 (Existence and uniqueness of polar form)

Hint 1. For which exponents

𝑛

did we prove Theorem B.1.20? Was

𝑧^{1 / 2}

ever defined?

Hint 2. What does de Moivre’s formula with

𝑛 = 2

say about

cos (𝜃 / 2) +

𝑖 sin (𝜃 / 2)

Sketch. Theorem B.1.20 holds for integer

𝑛

only, and

1^{1 / 2}

has no meaning:

1

has two square roots,

\pm 1

. The formula

cos (𝜃 / 2) +

𝑖 sin (𝜃 / 2)

depends on the argument

𝜃

chosen, not only on the number

𝑒^{𝑖 𝜃}

, and the arguments

0

and

2 𝜋

1

produce its two square roots.

The first sentence is the error. Theorem B.1.20 was proved for integers $𝑛$ , by induction and by inverting, and it says nothing for $𝑛 = 1 / 2$ . Worse, the symbol $1^{1 / 2}$ was never defined: $1$ has two square roots, $1$ and $- 1$ (Proposition B.1.29), and no rule has been given for choosing one.

What de Moivre’s formula does give, with $𝑛 = 2$ , is ${(cos (𝜃 / 2) + 𝑖 sin (𝜃 / 2))}^{2} =$ $cos 𝜃 +$ $𝑖 sin 𝜃$ . So $cos (𝜃 / 2) +$ $𝑖 sin (𝜃 / 2)$ is a square root of $𝑒^{𝑖 𝜃}$ . But it depends on $𝜃$ , not only on the number $𝑒^{𝑖 𝜃}$ , and a number has many arguments (Proposition B.1.15). The arguments $0$ and $2 𝜋$ of $1$ give its two different square roots, $1$ and $- 1$ , exactly as in Theorem B.1.26 with $𝑛 = 2$ . The rest of the computation is correct; it shows only that $1$ and $- 1$ are both square roots of $1$ .

Exercise B.1.40.standard For $𝐳 =$ $(𝑧_{1}, \dots, 𝑧_{𝑛}) \in$ $ℂ^{𝑛}$ , write $𝐳 = (𝑧_{1}, \dots, 𝑧_{𝑛})$ . Let $𝐴 \in ℝ^{𝑚 \times 𝑛}$ .

(a)Show that $𝐴 𝐳 = 𝐴 𝐳$ for every $𝐳 \in ℂ^{𝑛}$ .
(b)Let $𝑚 = 𝑛$ , $𝜆 \in ℂ$ , and $𝐯 \in ℂ^{𝑛}$ with $𝐴 𝐯 = 𝜆 𝐯$ . Show that $𝐴 𝐯 = 𝜆 𝐯$ .
(c)Let $𝐛 \in ℝ^{𝑚}$ . Show that the set of solutions $𝐳 \in ℂ^{𝑛}$ of $𝐴 𝐳 = 𝐛$ is closed under conjugation, and that if it is nonempty, it contains a vector in $ℝ^{𝑛}$ .

Solution.

Uses. Proposition 2.3.11 (Entries of $𝐴 𝐱$ ), Proposition B.1.2 (Conjugation respects the field operations)

Techniques. Conjugate the equation (Technique B.1.3)

Hint 1. Compute the

𝑖

th entry of

𝐴 𝐳

with Proposition 2.3.11 and use that each

𝑎_{𝑖 𝑗}

is real.

Hint 2. For (b) and (c), conjugate the given equation and apply (a). For the real solution in (c), average a solution with its conjugate.

Sketch. (a)

\sum_{𝑗} 𝑎_{𝑖 𝑗} 𝑧_{𝑗} =

\sum_{𝑗} 𝑎_{𝑖 𝑗} 𝑧_{𝑗}

because

𝑎_{𝑖 𝑗} = 𝑎_{𝑖 𝑗}

. (b)

𝐴 𝐯 =

𝐴 𝐯 =

𝜆 𝐯 =

𝜆 𝐯

. (c)

𝐴 𝐳 = 𝐛 = 𝐛

, and then

(𝐳 + 𝐳) / 2

, which has real entries

Re 𝑧_{𝑗}

, is a solution.

(a) By Proposition 2.3.11 and Proposition B.1.2, the $𝑖$ th entry of $𝐴 𝐳$ is

\sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝑧_{𝑗} = \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝑧_{𝑗} = \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝑧_{𝑗},

since $𝑎_{𝑖 𝑗} = 𝑎_{𝑖 𝑗}$ for real $𝑎_{𝑖 𝑗}$ . This is the $𝑖$ th entry of $𝐴 𝐳$ .

(b) By (a), $𝐴 𝐯 =$ $𝐴 𝐯 =$ $𝜆 𝐯 =$ $𝜆 𝐯$ , where the last step conjugates each entry $𝜆 𝑣_{𝑗}$ .

(c) If $𝐴 𝐳 = 𝐛$ , then by (a), $𝐴 𝐳 = 𝐛 = 𝐛$ , since $𝐛$ is real. So $𝐳$ is also a solution. Then $𝐱 = \frac{1}{2} (𝐳 + 𝐳)$ satisfies $𝐴 𝐱 = \frac{1}{2} (𝐛 + 𝐛) = 𝐛$ , and its entries $(𝑧_{𝑗} + 𝑧_{𝑗}) / 2 =$ $Re 𝑧_{𝑗}$ are real (Proposition B.1.2(d)).

In the language of Section 7.2, (b) says that the nonreal eigenvalues of a real matrix come in conjugate pairs $𝜆, 𝜆$ , with conjugate eigenvectors.

Exercise B.1.41.standard

(a)Find $𝐴 \in ℝ^{2 \times 2}$ with $𝐴^{3} = - 𝐼_{2}$ but $𝐴 \neq - 𝐼_{2}$ .
(b)For each $𝑛 \geq 2$ , find $𝐴 \in ℝ^{2 \times 2}$ with $𝐴^{𝑛} = 𝐼_{2}$ but $𝐴^{𝑘} \neq 𝐼_{2}$ for $1 \leq 𝑘 < 𝑛$ .

Solution.

Uses. Proposition B.1.18 (Multiplication by $𝑤$ as a map of the plane), Theorem 2.4.2 (The matrix of a composite), Theorem B.1.25 (The $𝑛$ th roots of unity)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. By Proposition B.1.18, complex numbers act on

ℝ^{2}

through the matrices

𝐽_{𝑤} = {[𝑀_{𝑤}]}_{ℬ︀ ℬ︀}

. How does

𝐽_{𝑤} 𝐽_{𝑧}

compare with

𝐽_{𝑤 𝑧}

Hint 2.

𝐽_{𝑤}^{𝑘} = 𝐽_{𝑤^{𝑘}}

. So look for

𝑤 \in ℂ

with

𝑤^{3} = - 1

but

𝑤 \neq - 1

, and in (b), for an

𝑛

th root of unity whose smaller powers are not

1

Sketch.

𝑀_{𝑤} 𝑀_{𝑧} = 𝑀_{𝑤 𝑧}

, so

𝐽_{𝑤}^{𝑘} = 𝐽_{𝑤^{𝑘}}

, and

𝐽_{𝑤} = 𝐽_{𝑤^{'}}

only if

𝑤 = 𝑤^{'}

. (a)

𝑤 = 𝑒^{𝑖 𝜋 / 3}

has

𝑤^{3} = - 1

, giving

𝐴 =

(\begin{matrix} 1 / 2 & - \sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{matrix})

. (b)

𝑤 = 𝑒^{2 𝜋 𝑖 / 𝑛}

, the rotation by

2 𝜋 / 𝑛

For $𝑤 = 𝑐 + 𝑑 𝑖$ let $𝐽_{𝑤} =$ ${[𝑀_{𝑤}]}_{ℬ︀ ℬ︀} =$ $(\begin{matrix} 𝑐 & - 𝑑 \\ 𝑑 & 𝑐 \end{matrix})$ , as in Proposition B.1.18. Since $𝑀_{𝑤} \circ 𝑀_{𝑧} = 𝑀_{𝑤 𝑧}$ , Theorem 2.4.2 gives $𝐽_{𝑤} 𝐽_{𝑧} = 𝐽_{𝑤 𝑧}$ , and by induction $𝐽_{𝑤}^{𝑘} = 𝐽_{𝑤^{𝑘}}$ for $𝑘 \geq 1$ . Also $𝐽_{𝑤}$ determines $𝑤$ (its first column is $(𝑐, 𝑑)$ ), and $𝐽_{1} = 𝐼_{2}$ , $𝐽_{- 1} = - 𝐼_{2}$ .

(a) Let $𝑤 =$ $𝑒^{𝑖 𝜋 / 3} =$ $1 / 2 +$ $(\sqrt{3} / 2) 𝑖$ . Then $𝑤^{3} = 𝑒^{𝑖 𝜋} = - 1$ by Theorem B.1.20, so

𝐴 = 𝐽_{𝑤} = (\begin{matrix} 1 / 2 & - \sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{matrix})

satisfies $𝐴^{3} = 𝐽_{- 1} = - 𝐼_{2}$ , and $𝐴 \neq - 𝐼_{2}$ . Geometrically, $𝐴$ is the rotation by $𝜋 / 3$ , and three of them make a half turn.

(b) Let $𝜔 = 𝑒^{2 𝜋 𝑖 / 𝑛}$ and $𝐴 =$ $𝐽_{𝜔} =$ $(\begin{matrix} cos (2 𝜋 / 𝑛) & - sin (2 𝜋 / 𝑛) \\ sin (2 𝜋 / 𝑛) & cos (2 𝜋 / 𝑛) \end{matrix})$ , the rotation by $2 𝜋 / 𝑛$ . Then $𝐴^{𝑘} = 𝐽_{𝜔^{𝑘}}$ , which equals $𝐼_{2} = 𝐽_{1}$ if and only if $𝜔^{𝑘} = 1$ , that is, $𝑛 ∣ 𝑘$ (Theorem B.1.25(a)). So $𝐴^{𝑛} = 𝐼_{2}$ and $𝐴^{𝑘} \neq 𝐼_{2}$ for $1 \leq 𝑘 < 𝑛$ .

Exercise B.1.42.standard For $𝑧 = 𝑎 + 𝑏 𝑖 \in ℂ$ , let $𝜑 (𝑧) =$ $(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) \in$ $ℝ^{2 \times 2}$ , and let $𝐾 = {𝜑 (𝑧) | 𝑧 \in ℂ}$ .

(a)Show that $𝜑 (𝑧 + 𝑤) =$ $𝜑 (𝑧) +$ $𝜑 (𝑤)$ , $𝜑 (𝑧 𝑤) =$ $𝜑 (𝑧) 𝜑 (𝑤)$ , and $𝜑 (1) = 𝐼_{2}$ , and that $𝜑$ is injective.
(b)Show that $𝜑 (𝑧) = {𝜑 (𝑧)}^{𝖳}$ and $det 𝜑 (𝑧) = {| 𝑧 |}^{2}$ .
(c)Show that $𝐾$ , with matrix addition and multiplication, is a field, and that ${𝜑 (𝑧)}^{- 1} = 𝜑 (𝑧^{- 1})$ for $𝑧 \neq 0$ .

Solution.

Uses. Definition 2.4.1 (Matrix product), Proposition 2.4.14 (Algebra of matrices), Definition 2.4.21 (Transpose), Definition 5.1.14 (The $2 \times 2$ determinant), Proposition B.1.18 (Multiplication by $𝑤$ as a map of the plane)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. Compute

𝜑 (𝑧) 𝜑 (𝑤)

directly, or note that

𝜑 (𝑧) = {[𝑀_{𝑧}]}_{ℬ︀ ℬ︀}

and

𝑀_{𝑧} \circ 𝑀_{𝑤} = 𝑀_{𝑧 𝑤}

Hint 2. For (c), most field axioms hold because they hold for all matrices (Proposition 2.4.14). Commutativity of multiplication and inverses can be transported from

ℂ

with (a).

Sketch. (a) Entrywise for sums;

(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) (\begin{matrix} 𝑐 & - 𝑑 \\ 𝑑 & 𝑐 \end{matrix}) =

(\begin{matrix} 𝑎 𝑐 - 𝑏 𝑑 & - (𝑎 𝑑 + 𝑏 𝑐) \\ 𝑎 𝑑 + 𝑏 𝑐 & 𝑎 𝑐 - 𝑏 𝑑 \end{matrix}) =

𝜑 (𝑧 𝑤)

. (b) Swap

𝑏

and

- 𝑏

;

det = 𝑎^{2} + 𝑏^{2}

. (c)

𝐾

contains

0

and

𝐼_{2}

and is closed under sums, negatives, and products;

𝜑 (𝑧) 𝜑 (𝑤) =

𝜑 (𝑧 𝑤) =

𝜑 (𝑤) 𝜑 (𝑧)

; and

𝜑 (𝑧) 𝜑 (𝑧^{- 1}) =

𝜑 (1) =

𝐼_{2}

Let $𝑧 = 𝑎 + 𝑏 𝑖$ and $𝑤 = 𝑐 + 𝑑 𝑖$ .

(a) $𝜑 (𝑧 + 𝑤) =$ $(\begin{matrix} 𝑎 + 𝑐 & - (𝑏 + 𝑑) \\ 𝑏 + 𝑑 & 𝑎 + 𝑐 \end{matrix}) =$ $𝜑 (𝑧) +$ $𝜑 (𝑤)$ . By Definition 2.4.1,

𝜑 (𝑧) 𝜑 (𝑤) = (\begin{matrix} 𝑎 𝑐 - 𝑏 𝑑 & - 𝑎 𝑑 - 𝑏 𝑐 \\ 𝑏 𝑐 + 𝑎 𝑑 & - 𝑏 𝑑 + 𝑎 𝑐 \end{matrix}) = 𝜑 ((𝑎 𝑐 - 𝑏 𝑑) + (𝑎 𝑑 + 𝑏 𝑐) 𝑖) = 𝜑 (𝑧 𝑤) .

(Equivalently, $𝜑 (𝑧) = {[𝑀_{𝑧}]}_{ℬ︀ ℬ︀}$ by Proposition B.1.18, and $𝑀_{𝑧} \circ 𝑀_{𝑤} = 𝑀_{𝑧 𝑤}$ .) Also $𝜑 (1) = 𝐼_{2}$ , and $𝜑 (𝑧)$ determines $𝑧$ through its first column $(𝑎, 𝑏)$ , so $𝜑$ is injective.

(b) $𝜑 (𝑧) =$ $(\begin{matrix} 𝑎 & 𝑏 \\ - 𝑏 & 𝑎 \end{matrix}) =$ ${𝜑 (𝑧)}^{𝖳}$ , and $det 𝜑 (𝑧) =$ $𝑎 \cdot$ $𝑎 -$ $(- 𝑏) \cdot$ $𝑏 =$ $𝑎^{2} +$ $𝑏^{2} =$ ${| 𝑧 |}^{2}$ (Definition 5.1.14).

(c) By (a), $𝐾$ is closed under addition and multiplication, and it contains $𝜑 (0) = 0$ , $𝜑 (1) = 𝐼_{2}$ , and the negative $- 𝜑 (𝑧) = 𝜑 (- 𝑧)$ of each of its elements. Addition of matrices is commutative and associative, and multiplication is associative and distributes over addition (Proposition 2.4.14), so these laws hold in $𝐾$ ; $0$ and $𝐼_{2}$ are identities, and $𝐼_{2} \neq 0$ . Multiplication in $𝐾$ is commutative, because $𝜑 (𝑧) 𝜑 (𝑤) =$ $𝜑 (𝑧 𝑤) =$ $𝜑 (𝑤 𝑧) =$ $𝜑 (𝑤) 𝜑 (𝑧)$ . Finally, a nonzero element of $𝐾$ is $𝜑 (𝑧)$ with $𝑧 \neq 0$ , because $𝜑$ is injective and $𝜑 (0) = 0$ ; then $𝜑 (𝑧) 𝜑 (𝑧^{- 1}) =$ $𝜑 (1) =$ $𝐼_{2} =$ $𝜑 (𝑧^{- 1}) 𝜑 (𝑧)$ , so $𝜑 (𝑧^{- 1}) \in 𝐾$ is its inverse. So $𝐾$ is a field, and $𝜑 : ℂ \to 𝐾$ is a bijection that respects sums and products: $𝐾$ is a copy of $ℂ$ made of real matrices, in which $𝑖$ is the quarter turn $𝜑 (𝑖) = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ , and in which conjugation is transposition.

Exercise B.1.43.standard Let

𝜎 : ℂ \to ℂ

be an automorphism of

ℂ

(Section B.1.1) with

𝜎 (𝑡) = 𝑡

for every

𝑡 \in ℝ

. Show that

𝜎

is either the identity or conjugation.

Solution.

Uses. Exercise 1.1.44, Proposition B.1.2 (Conjugation respects the field operations)

Hint 1.

𝜎

is determined by

𝜎 (𝑖)

, since

𝜎 (𝑎 + 𝑏 𝑖) =

𝜎 (𝑎) +

𝜎 (𝑏) 𝜎 (𝑖)

Hint 2. Apply

𝜎

𝑖^{2} = - 1

. Which complex numbers square to

- 1

Sketch.

{𝜎 (𝑖)}^{2} =

𝜎 (𝑖^{2}) =

𝜎 (- 1) =

- 1

, so

𝜎 (𝑖) = \pm 𝑖

. Then

𝜎 (𝑎 + 𝑏 𝑖) =

𝑎 +

𝑏 𝜎 (𝑖)

𝑎 + 𝑏 𝑖

𝑎 - 𝑏 𝑖

Since $𝜎$ respects products and fixes $- 1$ , ${𝜎 (𝑖)}^{2} =$ $𝜎 (𝑖^{2}) =$ $𝜎 (- 1) =$ $- 1$ . By Exercise 1.1.44(a), $𝜎 (𝑖) = 𝑖$ or $𝜎 (𝑖) = - 𝑖$ . For $𝑎, 𝑏 \in ℝ$ , since $𝜎$ respects sums and products and fixes $𝑎$ and $𝑏$ ,

𝜎 (𝑎 + 𝑏 𝑖) = 𝜎 (𝑎) + 𝜎 (𝑏) 𝜎 (𝑖) = 𝑎 + 𝑏 𝜎 (𝑖) .

If $𝜎 (𝑖) = 𝑖$ , this is $𝑎 + 𝑏 𝑖$ , so $𝜎$ is the identity; if $𝜎 (𝑖) = - 𝑖$ , it is $𝑎 - 𝑏 𝑖$ , so $𝜎$ is conjugation. Both are indeed automorphisms fixing $ℝ$ (Proposition B.1.2).

The hypothesis that $𝜎$ fixes $ℝ$ matters. Every automorphism fixes $ℚ$ (it fixes $1$ , hence every integer and every quotient of integers), but $ℂ$ has other, very discontinuous, automorphisms that move real numbers; proving that they exist needs the axiom of choice, for instance in the form of Zorn’s lemma (Section A.4).

Exercise B.1.44.standard Let

𝑧_{1}, 𝑧_{2}, 𝑧_{3} \in ℂ

with

| 𝑧_{1} | =

| 𝑧_{2} | =

| 𝑧_{3} | =

1

and

𝑧_{1} + 𝑧_{2} + 𝑧_{3} = 0

. Show that

{𝑧_{1}, 𝑧_{2}, 𝑧_{3}} =

{𝑢, 𝑢 𝜔, 𝑢 𝜔^{2}}

for some

𝑢

with

| 𝑢 | = 1

, where

𝜔 = 𝑒^{2 𝜋 𝑖 / 3}

. Conclude that

𝑧_{1}, 𝑧_{2}, 𝑧_{3}

are the vertices of an equilateral triangle.

Solution.

Uses. Proposition B.1.8 (Properties of the modulus), Proposition B.1.28 (Sum and product of the roots of unity), Theorem B.1.25 (The $𝑛$ th roots of unity)

Techniques. Rotate the phase (Technique 6.2.8), Square the modulus (Technique B.1.9)

Hint 1. Rotate so that

𝑧_{1} = 1

: divide everything by

𝑧_{1}

. The hypotheses are unchanged.

Hint 2. With

𝑤_{𝑘} = 𝑧_{𝑘} / 𝑧_{1}

, you have

𝑤_{3} = - 1 - 𝑤_{2}

. Take the squared modulus of both sides to find

Re 𝑤_{2}

Sketch. Let

𝑤_{𝑘} = 𝑧_{𝑘} / 𝑧_{1}

: then

𝑤_{1} = 1

| 𝑤_{2} | = | 𝑤_{3} | = 1

, and

1 + 𝑤_{2} + 𝑤_{3} = 0

. Now

1 =

{| 𝑤_{3} |}^{2} =

{| 1 + 𝑤_{2} |}^{2} =

2 +

2 Re 𝑤_{2}

, so

Re 𝑤_{2} = - 1 / 2

and

𝑤_{2} \in {𝜔, 𝜔^{2}}

; then

𝑤_{3} = - 1 - 𝑤_{2}

is the other one. So

𝑧_{𝑘} \in

{𝑧_{1}, 𝑧_{1} 𝜔, 𝑧_{1} 𝜔^{2}}

Since $| 𝑧_{1} | = 1$ , $𝑧_{1} \neq 0$ . Let $𝑤_{𝑘} = 𝑧_{𝑘} / 𝑧_{1}$ . Then $𝑤_{1} = 1$ , $| 𝑤_{2} | = | 𝑤_{3} | = 1$ by Proposition B.1.8(d), and dividing $𝑧_{1} + 𝑧_{2} + 𝑧_{3} = 0$ by $𝑧_{1}$ gives $1 + 𝑤_{2} + 𝑤_{3} = 0$ . So $𝑤_{3} = - (1 + 𝑤_{2})$ , and by Proposition B.1.8(f),

1 = {| 𝑤_{3} |}^{2} = {| 1 + 𝑤_{2} |}^{2} = 1 + 2 Re 𝑤_{2} + {| 𝑤_{2} |}^{2} = 2 + 2 Re 𝑤_{2} .

Hence $Re 𝑤_{2} = - 1 / 2$ , and ${(Im 𝑤_{2})}^{2} =$ $1 -$ $1 / 4 =$ $3 / 4$ , so $𝑤_{2} = - 1 / 2 \pm (\sqrt{3} / 2) 𝑖$ . These two numbers are $𝜔$ and $𝜔^{2} = 𝜔$ . By Proposition B.1.28, $1 + 𝜔 + 𝜔^{2} = 0$ , so if $𝑤_{2} = 𝜔$ then $𝑤_{3} = - 1 - 𝜔 = 𝜔^{2}$ , and if $𝑤_{2} = 𝜔^{2}$ then $𝑤_{3} = 𝜔$ . Either way ${𝑤_{1}, 𝑤_{2}, 𝑤_{3}} =$ ${1, 𝜔, 𝜔^{2}}$ , and multiplying by $𝑢 = 𝑧_{1}$ gives ${𝑧_{1}, 𝑧_{2}, 𝑧_{3}} =$ ${𝑢, 𝑢 𝜔, 𝑢 𝜔^{2}}$ .

For $𝑗 \neq 𝑘$ in ${0, 1, 2}$ , $| 𝑢 𝜔^{𝑗} -$ $𝑢 𝜔^{𝑘} | =$ $| 𝜔^{𝑗} | | 1 -$ $𝜔^{𝑘 - 𝑗} | =$ $| 1 -$ $𝜔^{𝑘 - 𝑗} |$ , and $𝜔^{𝑘 - 𝑗} \in$ ${𝜔, 𝜔^{2} = 𝜔}$ , so this is $| 1 -$ $𝜔 | =$ $| 3 / 2 -$ $(\sqrt{3} / 2) 𝑖 | =$ $\sqrt{3}$ in every case (Proposition B.1.8(b)). So all three sides have length $\sqrt{3}$ .

Exercise B.1.45.hard Let $𝑛 \geq 2$ and $𝜔 = 𝑒^{2 𝜋 𝑖 / 𝑛}$ .

(a)Show that $1 +$ $𝑥 +$ $\dots +$ $𝑥^{𝑛 - 1} =$ $(𝑥 - 𝜔) (𝑥 - 𝜔^{2}) \dots (𝑥 - 𝜔^{𝑛 - 1})$ in $ℂ [𝑥]$ .
(b)Deduce that $(1 - 𝜔) (1 - 𝜔^{2}) \dots (1 - 𝜔^{𝑛 - 1}) =$ $𝑛$ .
(c)Show that the product of the distances from one vertex of a regular $𝑛$ -gon inscribed in the unit circle to the other $𝑛 - 1$ vertices is $𝑛$ . Deduce that
$\prod_{𝑘 = 1}^{𝑛 - 1} sin \frac{𝑘 𝜋}{𝑛} = \frac{𝑛}{2^{𝑛 - 1}} .$

Solution.

Uses. Proposition B.1.28 (Sum and product of the roots of unity), Proposition 7.1.22 (Counting roots), Proposition B.1.8 (Properties of the modulus), Remark B.1.13 (Facts about $cos$ and $sin$ )

Techniques. Polynomial identity arguments (Technique 5.5.24)

Hint 1. Approach 2 of Proposition B.1.28 shows

𝑥^{𝑛} -

1 =

(𝑥 - 1) (𝑥 - 𝜔) \dots (𝑥 - 𝜔^{𝑛 - 1})

. Also

𝑥^{𝑛} -

1 =

(𝑥 - 1) (1 + 𝑥 + \dots + 𝑥^{𝑛 - 1})

. You cannot simply cancel

𝑥 - 1

yet; compare values instead.

Hint 2. Both sides of (a) take the same value at every

𝑐 \neq 1

. Two polynomials of degree at most

𝑛 - 1

that agree at infinitely many points are equal (Proposition 7.1.22).

Hint 3. For (c),

{| 1 - 𝑒^{𝑖 𝜃} |}^{2} =

2 -

2 cos 𝜃 =

4 {sin}^{2} (𝜃 / 2)

Sketch. Let

𝑃 = 1 + 𝑥 + \dots + 𝑥^{𝑛 - 1}

and

𝑄 =

\prod_{𝑘 = 1}^{𝑛 - 1} (𝑥 - 𝜔^{𝑘})

. For every

𝑐 \neq 1

(𝑐 - 1) 𝑃 (𝑐) =

𝑐^{𝑛} -

1 =

(𝑐 - 1) 𝑄 (𝑐)

, so

𝑃 (𝑐) = 𝑄 (𝑐)

; as

𝑃 - 𝑄

has degree at most

𝑛 - 1

and infinitely many roots,

𝑃 = 𝑄

. Evaluating at

1

gives (b). Taking moduli,

\prod | 1 - 𝜔^{𝑘} | = 𝑛

, and

| 1 -

𝜔^{𝑘} | =

2 sin (𝑘 𝜋 / 𝑛)

(a) Let $𝑃 = 1 + 𝑥 + \dots + 𝑥^{𝑛 - 1}$ and $𝑄 =$ $(𝑥 - 𝜔) (𝑥 - 𝜔^{2}) \dots (𝑥 - 𝜔^{𝑛 - 1})$ . By Approach 2 of Proposition B.1.28, $𝑥^{𝑛} - 1 = (𝑥 - 1) 𝑄$ , and multiplying out, $𝑥^{𝑛} - 1 = (𝑥 - 1) 𝑃$ . Evaluating at $𝑐 \in ℂ$ (evaluation respects products, Section 7.1),

(𝑐 - 1) 𝑃 (𝑐) = 𝑐^{𝑛} - 1 = (𝑐 - 1) 𝑄 (𝑐) .

For $𝑐 \neq 1$ we may cancel $𝑐 - 1$ , so $𝑃 (𝑐) = 𝑄 (𝑐)$ for every $𝑐 \neq 1$ . Both $𝑃$ and $𝑄$ have degree $𝑛 - 1$ , so $𝑃 - 𝑄$ has degree at most $𝑛 - 1$ and infinitely many roots, and $𝑃 - 𝑄 = 𝟎$ by Proposition 7.1.22.

(b) Evaluating (a) at $1$ gives $𝑛 =$ $𝑃 (1) =$ $𝑄 (1) =$ $(1 - 𝜔) (1 - 𝜔^{2}) \dots (1 - 𝜔^{𝑛 - 1})$ .

(c) The vertices of a regular $𝑛$ -gon inscribed in the unit circle are equally spaced on it, so they are $𝑢,$ $𝑢 𝜔,$ $\dots,$ $𝑢 𝜔^{𝑛 - 1}$ , where $𝑢$ is the chosen vertex and $| 𝑢 | = 1$ . Multiplication by $𝑢$ preserves distances, since $| 𝑢 𝑎 -$ $𝑢 𝑏 | =$ $| 𝑢 | | 𝑎 -$ $𝑏 | =$ $| 𝑎 -$ $𝑏 |$ , and it carries the vertices to $1, 𝜔, \dots, 𝜔^{𝑛 - 1}$ , with the chosen vertex going to $1$ . Taking moduli in (b), with Proposition B.1.8(d), the product of the distances is $\prod_{𝑘 = 1}^{𝑛 - 1} | 1 -$ $𝜔^{𝑘} | =$ $𝑛$ .

For $𝜃 \in ℝ$ , by (T1),

{| 1 - 𝑒^{𝑖 𝜃} |}^{2} = {(1 - cos 𝜃)}^{2} + {sin}^{2} 𝜃 = 2 - 2 cos 𝜃 = 4 {sin}^{2} (\frac{𝜃}{2}),

where the last step is $cos 𝜃 =$ ${cos}^{2} (𝜃 / 2) -$ ${sin}^{2} (𝜃 / 2) =$ $1 -$ $2 {sin}^{2} (𝜃 / 2)$ . For $𝜃 = 2 𝜋 𝑘 / 𝑛$ with $1 \leq 𝑘 \leq 𝑛 - 1$ , we have $0 < 𝜃 / 2 < 𝜋$ , so $sin (𝜃 / 2) > 0$ and $| 1 -$ $𝜔^{𝑘} | =$ $2 sin (𝑘 𝜋 / 𝑛)$ . Hence $2^{𝑛 - 1} \prod_{𝑘 = 1}^{𝑛 - 1} sin (𝑘 𝜋 / 𝑛) =$ $𝑛$ .

Exercise B.1.46.standard An $𝑛$ th root of unity $𝜁$ is primitive if $𝜁^{𝑚} \neq 1$ for $1 \leq 𝑚 < 𝑛$ . Let $𝜔 = 𝑒^{2 𝜋 𝑖 / 𝑛}$ .

(a)Show that for $0 \leq 𝑘 \leq 𝑛 - 1$ , $𝜔^{𝑘}$ is primitive if and only if $gcd (𝑘, 𝑛) = 1$ .
(b)Show that if $𝜁$ is primitive, then $1, 𝜁, 𝜁^{2}, \dots, 𝜁^{𝑛 - 1}$ are all the $𝑛$ th roots of unity.
(c)List the primitive $12$ th roots of unity in the form $𝑎 + 𝑏 𝑖$ .

Solution.

Uses. Theorem B.1.25 (The $𝑛$ th roots of unity), Euclid’s lemma: if $𝑛 ∣ 𝑘 𝑚$ and $gcd (𝑘, 𝑛) = 1$ , then $𝑛 ∣ 𝑚$

Hint 1. By Theorem B.1.25(a),

{(𝜔^{𝑘})}^{𝑚} = 1

if and only if

𝑛 ∣ 𝑘 𝑚

Hint 2. If

𝑔 = gcd (𝑘, 𝑛) > 1

, try

𝑚 = 𝑛 / 𝑔

. If

𝑔 = 1

, use Euclid’s lemma.

Hint 3. For (b), show that the

𝑛

powers are distinct, and count.

Sketch. (a)

{(𝜔^{𝑘})}^{𝑚} = 1

if and only if

𝑛 ∣ 𝑘 𝑚

. If

𝑔 = gcd (𝑘, 𝑛) > 1

, then

𝑚 = 𝑛 / 𝑔 < 𝑛

works, so

𝜔^{𝑘}

is not primitive. If

𝑔 = 1

, then

𝑛 ∣ 𝑘 𝑚

forces

𝑛 ∣ 𝑚

, so

𝑚 \geq 𝑛

. (b) If

𝜁^{𝑗} = 𝜁^{𝑙}

with

0 \leq 𝑗 < 𝑙 < 𝑛

, then

𝜁^{𝑙 - 𝑗} = 1

, which is impossible; so the powers are

𝑛

distinct

𝑛

th roots of unity. (c)

𝑘 = 1, 5, 7, 11

\pm \sqrt{3} / 2 \pm 𝑖 / 2

(a) By Theorem B.1.25(a), ${(𝜔^{𝑘})}^{𝑚} = 𝜔^{𝑘 𝑚} = 1$ if and only if $𝑛 ∣ 𝑘 𝑚$ . Let $𝑔 = gcd (𝑘, 𝑛)$ . If $𝑔 > 1$ , then $𝑚 = 𝑛 / 𝑔$ satisfies $1 \leq 𝑚 < 𝑛$ and $𝑘 𝑚 = (𝑘 / 𝑔) 𝑛$ , so ${(𝜔^{𝑘})}^{𝑚} = 1$ and $𝜔^{𝑘}$ is not primitive. If $𝑔 = 1$ and ${(𝜔^{𝑘})}^{𝑚} = 1$ with $𝑚 \geq 1$ , then $𝑛 ∣ 𝑘 𝑚$ , so $𝑛 ∣ 𝑚$ by Euclid’s lemma, and $𝑚 \geq 𝑛$ . So $𝜔^{𝑘}$ is primitive.

(b) Each $𝜁^{𝑗}$ is an $𝑛$ th root of unity, by Theorem B.1.25(c). If $𝜁^{𝑗} = 𝜁^{𝑙}$ with $0 \leq 𝑗 < 𝑙 \leq 𝑛 - 1$ , then $𝜁^{𝑙 - 𝑗} = 1$ with $1 \leq 𝑙 - 𝑗 < 𝑛$ , contradicting primitivity. So $1, 𝜁, \dots, 𝜁^{𝑛 - 1}$ are $𝑛$ distinct $𝑛$ th roots of unity, and by Theorem B.1.25(b) there are only $𝑛$ , so these are all of them.

(c) For $𝑛 = 12$ , the $𝑘$ in ${0, \dots, 11}$ with $gcd (𝑘, 12) = 1$ are $1, 5, 7, 11$ . So the primitive $12$ th roots of unity are

𝑒^{𝑖 𝜋 / 6} = \frac{\sqrt{3}}{2} + \frac{𝑖}{2}, 𝑒^{5 𝜋 𝑖 / 6} = - \frac{\sqrt{3}}{2} + \frac{𝑖}{2}, 𝑒^{7 𝜋 𝑖 / 6} = - \frac{\sqrt{3}}{2} - \frac{𝑖}{2}, 𝑒^{11 𝜋 𝑖 / 6} = \frac{\sqrt{3}}{2} - \frac{𝑖}{2} .

Exercise B.1.47.hard Let

𝑛 \geq 1

. Show that

𝑥^{2} + 𝑥 + 1

divides

𝑥^{2 𝑛} + 𝑥^{𝑛} + 1

ℝ [𝑥]

if and only if

3 ∤ 𝑛

Solution.

Uses. Theorem 7.1.14 (Division with remainder), Proposition B.1.28 (Sum and product of the roots of unity), Theorem B.1.25 (The $𝑛$ th roots of unity)

Hint 1. Let

𝑝 = 𝑥^{2 𝑛} + 𝑥^{𝑛} + 1

and

𝜔 = 𝑒^{2 𝜋 𝑖 / 3}

, a root of

𝑥^{2} + 𝑥 + 1

. Divide

𝑝

𝑥^{2} + 𝑥 + 1

with remainder in

ℝ [𝑥]

, and evaluate at

𝜔

Hint 2. The remainder is

𝑐 𝑥 + 𝑑

with

𝑐, 𝑑 \in ℝ

, and

𝑝 (𝜔) = 𝑐 𝜔 + 𝑑

. Since

𝜔

is not real,

𝑐 𝜔 + 𝑑 = 0

forces

𝑐 = 𝑑 = 0

Hint 3. Compute

𝑝 (𝜔) = 1 + 𝜁 + 𝜁^{2}

with

𝜁 = 𝜔^{𝑛}

. When is

𝜁 = 1

Sketch. Write

𝑝 =

𝑞 (𝑥^{2} + 𝑥 + 1) +

(𝑐 𝑥 + 𝑑)

with

𝑞 \in ℝ [𝑥]

and

𝑐, 𝑑 \in ℝ

. Since

𝜔^{2} + 𝜔 + 1 = 0

𝑝 (𝜔) = 𝑐 𝜔 + 𝑑

, whose imaginary part is

𝑐 \sqrt{3} / 2

; so the remainder vanishes if and only if

𝑝 (𝜔) = 0

. With

𝜁 = 𝜔^{𝑛}

𝑝 (𝜔) = 1 + 𝜁 + 𝜁^{2}

, which is

3

3 ∣ 𝑛

and

0

otherwise, since then

𝜁

is a cube root of unity other than

1

Let $𝑝 = 𝑥^{2 𝑛} + 𝑥^{𝑛} + 1$ and $𝜔 =$ $𝑒^{2 𝜋 𝑖 / 3} =$ $- 1 / 2 +$ $(\sqrt{3} / 2) 𝑖$ . By Proposition B.1.28, $𝜔^{2} + 𝜔 + 1 = 0$ . By the division algorithm in $ℝ [𝑥]$ (Theorem 7.1.14),

𝑝 = 𝑞 (𝑥^{2} + 𝑥 + 1) + (𝑐 𝑥 + 𝑑)

for some $𝑞 \in ℝ [𝑥]$ and $𝑐, 𝑑 \in ℝ$ , and $𝑥^{2} + 𝑥 + 1$ divides $𝑝$ if and only if $𝑐 = 𝑑 = 0$ . Evaluating at $𝜔$ (evaluation respects sums and products, Section 7.1) gives $𝑝 (𝜔) =$ $𝑞 (𝜔) \cdot$ $0 +$ $𝑐 𝜔 +$ $𝑑 =$ $𝑐 𝜔 +$ $𝑑$ . Now $𝑐 𝜔 +$ $𝑑 =$ $(𝑑 - 𝑐 / 2) +$ $(𝑐 \sqrt{3} / 2) 𝑖$ is $0$ if and only if $𝑐 = 0$ and $𝑑 = 0$ . So $𝑥^{2} + 𝑥 + 1$ divides $𝑝$ if and only if $𝑝 (𝜔) = 0$ .

Let $𝜁 = 𝜔^{𝑛}$ . Then $𝑝 (𝜔) = 1 + 𝜁 + 𝜁^{2}$ , and $𝜁^{3} = {(𝜔^{3})}^{𝑛} = 1$ . If $3 ∣ 𝑛$ , then $𝜁 = 1$ by Theorem B.1.25(a), and $𝑝 (𝜔) = 3 \neq 0$ . If $3 ∤ 𝑛$ , then $𝜁 \neq 1$ by Theorem B.1.25(a), so $(𝜁 - 1) (1 + 𝜁 + 𝜁^{2}) =$ $𝜁^{3} -$ $1 =$ $0$ gives $1 + 𝜁 + 𝜁^{2} = 0$ . So $𝑝 (𝜔) = 0$ if and only if $3 ∤ 𝑛$ .

For instance, $𝑥^{4} +$ $𝑥^{2} +$ $1 =$ $(𝑥^{2} + 𝑥 + 1) (𝑥^{2} - 𝑥 + 1)$ , while $𝑥^{6} + 𝑥^{3} + 1$ is not divisible by $𝑥^{2} + 𝑥 + 1$ .

Exercise B.1.48.standardBeyond 21-242 The Gaussian integers are $ℤ [𝑖] =$ ${𝑎 + 𝑏 𝑖 | 𝑎, 𝑏 \in ℤ}$ , and the norm of $𝑧 \in ℤ [𝑖]$ is $𝑁 (𝑧) = {| 𝑧 |}^{2}$ .

(a)Show that $ℤ [𝑖]$ is closed under addition, negation, and multiplication, that $𝑁 (𝑧)$ is a nonnegative integer, and that $𝑁 (𝑧 𝑤) =$ $𝑁 (𝑧) 𝑁 (𝑤)$ .
(b)A unit of $ℤ [𝑖]$ is a $𝑧 \in ℤ [𝑖]$ with $𝑧 𝑤 = 1$ for some $𝑤 \in ℤ [𝑖]$ . Show that the units are $1, - 1, 𝑖, - 𝑖$ .
(c)Starting from $5 = 𝑁 (2 + 𝑖)$ and $13 = 𝑁 (3 + 2 𝑖)$ , write $65$ as a sum of two squares of integers in two different ways.

Solution.

Uses. Proposition B.1.8 (Properties of the modulus), Definition 1.1.29 (Complex numbers)

Hint 1. For (a), use the product formula of Definition 1.1.29 and Proposition B.1.8(d).

Hint 2. For (b), take norms in

𝑧 𝑤 = 1

. Which integers

𝑎, 𝑏

have

𝑎^{2} + 𝑏^{2} = 1

Hint 3. For (c), compute

𝑁 ((2 + 𝑖) (3 + 2 𝑖))

and

𝑁 ((2 + 𝑖) (3 - 2 𝑖))

Sketch. (a)

(𝑎 + 𝑏 𝑖) (𝑐 + 𝑑 𝑖) =

(𝑎 𝑐 - 𝑏 𝑑) +

(𝑎 𝑑 + 𝑏 𝑐) 𝑖

has integer coordinates,

𝑁 (𝑎 + 𝑏 𝑖) =

𝑎^{2} +

𝑏^{2}

, and

𝑁 (𝑧 𝑤) =

{| 𝑧 𝑤 |}^{2} =

{| 𝑧 |}^{2} {| 𝑤 |}^{2}

. (b)

𝑁 (𝑧) 𝑁 (𝑤) = 1

forces

𝑁 (𝑧) = 1

, so

𝑧 \in {\pm 1, \pm 𝑖}

; each of these has an inverse in

ℤ [𝑖]

. (c)

(2 + 𝑖) (3 + 2 𝑖) =

4 +

7 𝑖

and

(2 + 𝑖) (3 - 2 𝑖) =

8 -

𝑖

, so

65 = 4^{2} + 7^{2} = 8^{2} + 1^{2}

(a) For $𝑧 = 𝑎 + 𝑏 𝑖$ and $𝑤 = 𝑐 + 𝑑 𝑖$ with integer coordinates, $𝑧 + 𝑤$ , $- 𝑧$ , and $𝑧 𝑤 =$ $(𝑎 𝑐 - 𝑏 𝑑) +$ $(𝑎 𝑑 + 𝑏 𝑐) 𝑖$ have integer coordinates. $𝑁 (𝑧) = 𝑎^{2} + 𝑏^{2}$ is a nonnegative integer, and $𝑁 (𝑧 𝑤) =$ ${| 𝑧 𝑤 |}^{2} =$ ${| 𝑧 |}^{2} {| 𝑤 |}^{2} =$ $𝑁 (𝑧) 𝑁 (𝑤)$ by Proposition B.1.8(d).

(b) If $𝑧 𝑤 = 1$ with $𝑧, 𝑤 \in ℤ [𝑖]$ , then $𝑁 (𝑧) 𝑁 (𝑤) =$ $𝑁 (1) =$ $1$ by (a), and $𝑁 (𝑧), 𝑁 (𝑤)$ are nonnegative integers, so $𝑁 (𝑧) = 1$ . If $𝑧 = 𝑎 + 𝑏 𝑖$ , then $𝑎^{2} + 𝑏^{2} = 1$ with $𝑎, 𝑏 \in ℤ$ , so one of $𝑎, 𝑏$ is $\pm 1$ and the other is $0$ : $𝑧 \in {1, - 1, 𝑖, - 𝑖}$ . Conversely, $1 \cdot$ $1 =$ $(- 1) (- 1) =$ $𝑖 (- 𝑖) =$ $1$ , so all four are units.

(c) By (a), $𝑁 ((2 + 𝑖) (3 + 2 𝑖)) =$ $5 \cdot$ $13 =$ $65$ , and $(2 + 𝑖) (3 + 2 𝑖) =$ $6 +$ $4 𝑖 +$ $3 𝑖 +$ $2 𝑖^{2} =$ $4 +$ $7 𝑖$ , so $65 = 4^{2} + 7^{2}$ . Also $𝑁 (3 - 2 𝑖) = 13$ , and $(2 + 𝑖) (3 - 2 𝑖) =$ $6 -$ $4 𝑖 +$ $3 𝑖 -$ $2 𝑖^{2} =$ $8 -$ $𝑖$ , so $65 = 8^{2} + 1^{2}$ . In coordinates, (a) is the identity $(𝑎^{2} + 𝑏^{2}) (𝑐^{2} + 𝑑^{2}) =$ ${(𝑎 𝑐 - 𝑏 𝑑)}^{2} +$ ${(𝑎 𝑑 + 𝑏 𝑐)}^{2}$ noted after Proposition B.1.8.

Exercise B.1.49.challenge Let $𝑧_{1}, \dots, 𝑧_{𝑛} \in ℂ$ . Show that there is a set $𝑆 \subseteq {1, \dots, 𝑛}$ with

| \sum_{𝑘 \in 𝑆} 𝑧_{𝑘} | \geq \frac{1}{𝜋} \sum_{𝑘 = 1}^{𝑛} | 𝑧_{𝑘} | .

Solution.

Uses. Proposition B.1.8 (Properties of the modulus), Notation B.1.19 ( $𝑒^{𝑖 𝜃}$ ), integrals of continuous functions

Techniques. Rotate the phase (Technique 6.2.8)

Hint 1. For a direction

𝑒^{𝑖 𝜃}

, keep the terms that make an acute angle with it:

𝑆_{𝜃} =

{𝑘 | Re (𝑒^{- 𝑖 𝜃} 𝑧_{𝑘}) > 0}

. Bound

| \sum_{𝑘 \in 𝑆_{𝜃}} 𝑧_{𝑘} |

from below by a real part.

Hint 2.

| \sum_{𝑘 \in 𝑆_{𝜃}} 𝑧_{𝑘} | \geq

𝑓 (𝜃) =

\sum_{𝑘} max (0, Re (𝑒^{- 𝑖 𝜃} 𝑧_{𝑘}))

. You cannot choose the best

𝜃

directly; average

𝑓

over

𝜃 \in [0, 2 𝜋]

instead.

Hint 3. With

𝑧_{𝑘} =

| 𝑧_{𝑘} | 𝑒^{𝑖 𝜑_{𝑘}}

Re (𝑒^{- 𝑖 𝜃} 𝑧_{𝑘}) =

| 𝑧_{𝑘} | cos (𝜃 - 𝜑_{𝑘})

, and the integral of

max (0, cos 𝑡)

over a period is

2

Sketch. For each

𝜃

| \sum_{𝑆_{𝜃}} 𝑧_{𝑘} | \geq

Re (𝑒^{- 𝑖 𝜃} \sum_{𝑆_{𝜃}} 𝑧_{𝑘}) =

𝑓 (𝜃)

. Since

\int_{0}^{2 𝜋} 𝑓 =

\sum_{𝑘} | 𝑧_{𝑘} | \int_{- 𝜋 / 2}^{𝜋 / 2} cos 𝑡 d 𝑡 =

2 \sum_{𝑘} | 𝑧_{𝑘} |

, the average of

𝑓

over

[0, 2 𝜋]

(1 / 𝜋) \sum_{𝑘} | 𝑧_{𝑘} |

, and the continuous function

𝑓

reaches its average somewhere.

For $𝜃 \in ℝ$ let $𝑆_{𝜃} =$ ${𝑘 | Re (𝑒^{- 𝑖 𝜃} 𝑧_{𝑘}) > 0}$ and

𝑓 (𝜃) = \sum_{𝑘 = 1}^{𝑛} max (0, Re (𝑒^{- 𝑖 𝜃} 𝑧_{𝑘})) = \sum_{𝑘 \in 𝑆_{𝜃}} Re (𝑒^{- 𝑖 𝜃} 𝑧_{𝑘}) .

By Proposition B.1.8(d) and (e),

| \sum_{𝑘 \in 𝑆_{𝜃}} 𝑧_{𝑘} | = | 𝑒^{- 𝑖 𝜃} \sum_{𝑘 \in 𝑆_{𝜃}} 𝑧_{𝑘} | \geq Re (\sum_{𝑘 \in 𝑆_{𝜃}} 𝑒^{- 𝑖 𝜃} 𝑧_{𝑘}) = 𝑓 (𝜃) .

So it suffices to find a $𝜃$ with $𝑓 (𝜃) \geq$ $(1 / 𝜋) \sum_{𝑘} | 𝑧_{𝑘} |$ .

Write $𝑧_{𝑘} =$ $| 𝑧_{𝑘} | 𝑒^{𝑖 𝜑_{𝑘}}$ , with any $𝜑_{𝑘}$ if $𝑧_{𝑘} = 0$ . Then $Re (𝑒^{- 𝑖 𝜃} 𝑧_{𝑘}) =$ $| 𝑧_{𝑘} | Re (𝑒^{𝑖 (𝜑_{𝑘} - 𝜃)}) =$ $| 𝑧_{𝑘} | cos (𝜃 - 𝜑_{𝑘})$ . So $𝑓$ is continuous and $2 𝜋$ -periodic, and integrating each term over a period centred at $𝜑_{𝑘}$ ,

\int_{0}^{2 𝜋} 𝑓 (𝜃) d 𝜃 = \sum_{𝑘 = 1}^{𝑛} | 𝑧_{𝑘} | \int_{𝜑_{𝑘} - 𝜋}^{𝜑_{𝑘} + 𝜋} max (0, cos (𝜃 - 𝜑_{𝑘})) d 𝜃 = \sum_{𝑘 = 1}^{𝑛} | 𝑧_{𝑘} | \int_{- 𝜋 / 2}^{𝜋 / 2} cos 𝑡 d 𝑡 = 2 \sum_{𝑘 = 1}^{𝑛} | 𝑧_{𝑘} | .

If we had $𝑓 (𝜃) <$ $(1 / 𝜋) \sum_{𝑘} | 𝑧_{𝑘} |$ for every $𝜃 \in [0, 2 𝜋]$ , then, as $𝑓$ is continuous, the integral would be less than $2 𝜋 \cdot$ $(1 / 𝜋) \sum_{𝑘} | 𝑧_{𝑘} | =$ $2 \sum_{𝑘} | 𝑧_{𝑘} |$ , a contradiction. So some $𝜃$ has $𝑓 (𝜃) \geq$ $(1 / 𝜋) \sum_{𝑘} | 𝑧_{𝑘} |$ , and $𝑆 = 𝑆_{𝜃}$ works.

This is an averaging argument. We cannot say in advance which direction $𝜃$ is best, but the average of $𝑓$ over all directions is computable, and some direction does at least as well as the average.

More practice

The practice bank has 16 exercises on this section’s topics.