---
title: "Complex Numbers"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 3
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/3-complex-numbers
---

# Chapter 3 — Complex Numbers

Complex numbers were met in the High School volume as a computational device for quadratic equations. This chapter treats them as a central object: the exponential form and its consequences (de Moivre, $n$-th roots, [roots of unity](#def-b1-complex-unity)), the systematic translation between $\C$ and plane geometry, and the use of $\eu^{\iu\theta}$ as a machine for proving trigonometric identities.

## 3.1 The field $\C$, modulus and conjugate

**Definition 3.1 (The complex field).**

$\C = \{a + \iu b : a, b \in \R\}$ with the usual addition and the multiplication determined by $\iu^2 = -1$. Every nonzero $z = a + \iu
b$ has an inverse: $z^{-1} = \frac{a - \iu b}{a^2 + b^2}$ — in the language of [Chapter 7](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#ch-b1-structures), $\C$ is a field. One writes $a =
\Re(z)$, $b = \Im(z)$, $\conj{z} = a - \iu b$ (the *conjugate*) and $\abs{z} = \sqrt{a^2 + b^2}$ (the *modulus*).

**Proposition 3.2 (Rules for conjugate and modulus).**

For $z, w \in \C$:

1. $\conj{z + w} = \conj z + \conj w$ , $\conj{zw} = \conj z\,  \conj w$ , $\conj{\conj z} = z$ ;
2. $z \conj z = \abs z^2$ ; $\;\Re(z) = \frac{z + \conj z}{2}$ , $\Im(z) = \frac{z - \conj z}{2\iu}$ ;
3. $\abs{zw} = \abs z\, \abs w$ , and $\abs{z^{-1}} =  \abs{z}^{-1}$ for $z \neq 0$ ;
4. (triangle inequality) $\abs{z + w}  \leq \abs z + \abs w$ , with equality if and only if $z$ and $w$ lie on a common ray from $0$ ( $w = \lambda z$ or $z = \lambda w$ with $\lambda \geq 0$ );
5. (reverse triangle inequality) $\bigl|\abs z - \abs w\bigr|  \leq \abs{z - w}$ .

**Proof.** (1) and (2) are direct computations on real and imaginary parts. (3): $\abs{zw}^2 = zw\,\conj{zw} = z\conj z\, w \conj w = \abs z^2 \abs
w^2$, then take square roots; apply to $z \cdot z^{-1} = 1$ for the inverse.

(4) Both sides are nonnegative, so compare squares:

$$
\abs{z+w}^2 = (z+w)(\conj z + \conj w)
= \abs z^2 + \abs w^2 + 2\,\Re(z \conj w),
$$

and $\Re(z\conj w) \leq \abs{z \conj w} = \abs z \abs w$ gives $\abs{z+w}^2 \leq (\abs z + \abs w)^2$. Equality forces $\Re(z \conj
w) = \abs{z \conj w}$, i.e. $z \conj w \in \R_+$; if $w \neq 0$ this gives $z = \frac{z\conj w}{\abs w^2}\, w = \lambda w$ with $\lambda
\geq 0$ (and the case $w = 0$ is trivial).

(5) $\abs z = \abs{(z - w) + w} \leq \abs{z-w} + \abs w$ gives $\abs z
- \abs w \leq \abs{z - w}$; swap $z$ and $w$ for the other sign. ∎

**Example 3.3 (A complete modulus-and-argument workout).**

Put $w = \dfrac{3 + 4\iu}{1 - 2\iu}$ in algebraic form and compute its [modulus](#def-b1-complex-field) twice. Multiplying by the [conjugate](#def-b1-complex-field) of the denominator:

$$
w = \frac{(3 + 4\iu)(1 + 2\iu)}{(1 - 2\iu)(1 + 2\iu)}
= \frac{3 + 6\iu + 4\iu - 8}{1 + 4}
= \frac{-5 + 10\iu}{5} = -1 + 2\iu .
$$

Directly: $\abs w = \sqrt{1 + 4} = \sqrt5$. Via the quotient rule ([Proposition 3.2](#prop-b1-complex-rules) (3)): $\abs w = \frac{\abs{3 +
4\iu}}{\abs{1 - 2\iu}} = \frac5{\sqrt5} = \sqrt5$ — same answer, no algebraic form needed. The lesson generalizes: moduli and arguments travel well through products and quotients, real and imaginary parts travel well through sums. Choose the representation that matches the operations at hand, and convert only when forced.

**Example 3.4 (Equations involving the conjugate).**

Solve in $\C$: $\;z + 2\conj z = 6 + 2\iu$. An equation mixing $z$ and $\conj z$ is *not* polynomial in $z$; the reliable move is to split into real coordinates. With $z = x + \iu y$:

$$
z + 2\conj z = 3x - \iu y ,
$$

so the equation reads $3x = 6$ and $-y = 2$: the unique solution is $z = 2 - 2\iu$. (Check: $(2 - 2\iu) + 2(2 + 2\iu) = 6 +
2\iu$.) Alternatively, [conjugate](#def-b1-complex-field) the whole equation to get $\conj z + 2z = 6 - 2\iu$ and solve the linear system in the unknowns $z, \conj z$ — same answer, and a useful trick when the coefficients are complex. Equations in $z$ and $\conj z$ are really systems of two real equations; expecting “degree $1$, one solution” is safe here, but $z\conj z = -1$ (no solution) shows the polynomial intuition failing as soon as products appear.

## 3.2 Exponential form

**Definition 3.5 (Complex exponential of an imaginary argument).**

For $\theta \in \R$ one defines

$$
\eu^{\iu\theta} = \cos\theta + \iu \sin\theta .
$$

Every $z \neq 0$ can be written $z = r\,\eu^{\iu\theta}$ with $r =
\abs z > 0$; $\theta$ is an *argument* of $z$, determined up to adding a multiple of $2\pi$. The value in $\intoc{-\pi}{\pi}$ is the *principal argument*, written $\arg z$.

**Example 3.6 (First values, and one famous identity).**

Reading the definition at the cardinal angles:

$$
\eu^{\iu\pi/2} = \iu, \qquad
\eu^{\iu\pi} = -1, \qquad
\eu^{2\iu\pi} = 1, \qquad
\eu^{\iu\pi/4} = \frac{\sqrt2}2\,(1 + \iu) .
$$

The second one, rearranged as $\eu^{\iu\pi} + 1 = 0$, is Euler’s celebrated identity linking $\eu$, $\iu$, $\pi$, $1$ and $0$; at this stage of the book it is a definition unwinding rather than a theorem, and its real content — why the analytic exponential function of [Chapter 4](https://one-course.com/books/math/3/en/chapter/4-standard-functions#ch-b1-functions), extended to imaginary arguments, deserves the same name — is settled by the power series of [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series). Meanwhile the display above is worth memorizing as a conversion table: it is used silently every time an argument is read off a picture.

**Theorem 3.7 (The functional equation).**

For all $\theta, \varphi \in \R$:

$$
\eu^{\iu\theta}\, \eu^{\iu\varphi} = \eu^{\iu(\theta + \varphi)},
\qquad
\abs{\eu^{\iu\theta}} = 1,
\qquad
\conj{\eu^{\iu\theta}} = \eu^{-\iu\theta} = (\eu^{\iu\theta})^{-1}.
$$

Consequently $\abs{zw} = \abs z \abs w$ carries the arguments too: $\arg(zw) \equiv \arg z + \arg w \pmod{2\pi}$.

**Proof.** Expand the product and use the addition formulas:

$$
(\cos\theta + \iu\sin\theta)(\cos\varphi + \iu\sin\varphi)
= (\cos\theta\cos\varphi - \sin\theta\sin\varphi)
+ \iu\,(\sin\theta\cos\varphi + \cos\theta\sin\varphi),
$$

which is $\cos(\theta+\varphi) + \iu\sin(\theta+\varphi)$. The [modulus](#def-b1-complex-field) is $\sqrt{\cos^2\theta + \sin^2\theta} = 1$, and the [conjugate](#def-b1-complex-field) formula is the parity of cosine and sine; it inverts $\eu^{\iu\theta}$ since $\eu^{\iu\theta}\eu^{-\iu\theta} = \eu^0 = 1$. ∎

**Corollary 3.8 (de Moivre’s formula).**

For $\theta \in \R$ and $n \in \Z$: $\;(\cos\theta + \iu\sin\theta)^n = \cos(n\theta) +
\iu\sin(n\theta)$.

**Proof.** For $n \geq 0$, induct. The case $n = 0$ reads $1 = 1$. Assuming the formula for $n$, the functional equation ([Theorem 3.7](#thm-b1-complex-funceq)) gives

$$
\bigl(\eu^{\iu\theta}\bigr)^{n+1}
= \bigl(\eu^{\iu\theta}\bigr)^{n}\,\eu^{\iu\theta}
= \eu^{\iu n\theta}\,\eu^{\iu\theta}
= \eu^{\iu(n+1)\theta} ,
$$

which is the formula at rank $n + 1$. For $n < 0$, write $n = -m$ with $m > 0$: since $\eu^{\iu m\theta}$ has inverse $\eu^{-\iu m\theta}$ ([Theorem 3.7](#thm-b1-complex-funceq) again),

$$
\bigl(\eu^{\iu\theta}\bigr)^{-m}
= \bigl(\eu^{\iu m\theta}\bigr)^{-1}
= \eu^{\iu(-m)\theta} . \qedhere
$$

∎

**Example 3.9 (Expanding cos⁡3θ\cos 3\thetacos3θ by de Moivre).**

Write $c = \cos\theta$, $s = \sin\theta$. De Moivre and the binomial theorem give

$$
\cos 3\theta + \iu \sin 3\theta = (c + \iu s)^3
= c^3 - 3cs^2 + \iu\,(3c^2 s - s^3),
$$

and identifying real parts, then substituting $s^2 = 1 - c^2$:

$$
\cos 3\theta = c^3 - 3c(1 - c^2) = 4\cos^3\theta - 3\cos\theta .
$$

The imaginary part yields $\sin 3\theta = 3\sin\theta -
4\sin^3\theta$ for free: one complex identity always carries *two* real ones. Read backwards, the boxed identity is the key to the classical trisection equation: constructing $\cos(\theta/3)$ from $\cos\theta$ means solving the cubic $4x^3 -
3x = \cos\theta$, which is where the algebra of [Chapter 8](https://one-course.com/books/math/3/en/chapter/8-polynomials#ch-b1-poly) takes over.

**Proposition 3.10 (Euler’s formulas).**

$$
\cos\theta = \frac{\eu^{\iu\theta} + \eu^{-\iu\theta}}{2},
\qquad
\sin\theta = \frac{\eu^{\iu\theta} - \eu^{-\iu\theta}}{2\iu}.
$$

**Proof.** Add, respectively subtract, $\eu^{\iu\theta} = \cos\theta +
\iu\sin\theta$ and $\eu^{-\iu\theta} = \cos\theta - \iu\sin\theta$. ∎

**Method 3.11 (Trigonometry via exponentials).**

1. *Linearize* $\cos^p\theta \sin^q\theta$ (turn powers into a sum of $\cos k\theta$ , $\sin k\theta$ ): substitute Euler’s formulas, expand with the binomial theorem ( [Theorem 2.16](https://one-course.com/books/math/3/en/chapter/2-counting#thm-b1-counting-binomial) ), regroup [conjugate](#def-b1-complex-field) terms.
2. *Expand* $\cos n\theta$ as a polynomial in $\cos\theta$ : write $\cos n\theta = \Re\bigl((\cos\theta +  \iu\sin\theta)^n\bigr)$ , expand, and convert even powers of $\sin$ via $\sin^2 = 1 - \cos^2$ .
3. *Sum trigonometric series* like $\sum_k \cos k\theta$ : recognize the real part of a geometric sum $\sum_k  (\eu^{\iu\theta})^k$ .
4. The *half-angle factorization*: for all $p, q$, $$\eu^{\iu p} + \eu^{\iu q}  = 2 \cos\tfrac{p - q}{2}\; \eu^{\iu \frac{p+q}{2}},  \qquad  \eu^{\iu p} - \eu^{\iu q}  = 2\iu \sin\tfrac{p - q}{2}\; \eu^{\iu \frac{p+q}{2}} .$$

**Example 3.12 (Linearization).**

$$
\cos^3\theta
= \Bigl(\frac{\eu^{\iu\theta} + \eu^{-\iu\theta}}{2}\Bigr)^{\!3}
= \frac{\eu^{3\iu\theta} + 3\eu^{\iu\theta} + 3\eu^{-\iu\theta} +
\eu^{-3\iu\theta}}{8}
= \frac{\cos 3\theta + 3\cos\theta}{4}.
$$

This form integrates immediately — the very reason linearization matters in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration). A mixed product works the same way, only with both Euler formulas at once:

$$
\sin^2\theta\cos^2\theta
= \Bigl(\frac{\sin2\theta}2\Bigr)^{\!2}
= \frac{1}{4}\cdot
\Bigl(\frac{\eu^{2\iu\theta} - \eu^{-2\iu\theta}}{2\iu}
\Bigr)^{\!2}
= \frac{2 - \eu^{4\iu\theta} - \eu^{-4\iu\theta}}{16}
= \frac{1 - \cos4\theta}{8} ,
$$

where the double-angle shortcut in the first step saved one binomial expansion — always worth scanning for before mechanizing.

**Example 3.13 (A binomial trigonometric sum).**

For $n \in \N$ and $\theta \in \R$, evaluate $S =
\sum_{k=0}^{n}\binom nk \cos k\theta$. Recognize the real part of a binomial expansion:

$$
S = \Re\sum_{k=0}^n \binom nk \bigl(\eu^{\iu\theta}\bigr)^k
= \Re\bigl(1 + \eu^{\iu\theta}\bigr)^n ,
$$

then factor the half-angle ([Method 3.11](#met-b1-complex-trig) (4)): $1 +
\eu^{\iu\theta} = 2\cos\frac\theta2\,\eu^{\iu\theta/2}$, so

$$
S = \Re\Bigl(2^n\cos^n\frac\theta2\;\eu^{\iu n\theta/2}\Bigr)
= 2^n \cos^n\frac\theta2\,\cos\frac{n\theta}2 .
$$

The imaginary part yields $\sum_k\binom nk\sin k\theta =
2^n\cos^n\frac\theta2\sin\frac{n\theta}2$ for free. Sanity checks: $\theta = 0$ recovers $\sum\binom nk = 2^n$, and $\theta = \pi$ gives $S = 0$ for $n \geq 1$ (each factor $\cos\frac\pi2$ vanishes), i.e. the alternating row sum of [Example 2.17](https://one-course.com/books/math/3/en/chapter/2-counting#ex-b1-counting-binomial). The method — “see the cosine sum as the shadow of a complex power, then factor half-angles” — is exactly that of [Exercise 3.6](#exo-b1-complex-6), with the binomial theorem replacing the geometric series.

## 3.3 Roots of complex numbers

**Theorem 3.14 (nnn-th roots).**

Let $a = r\,\eu^{\iu\alpha} \neq 0$ and $n \in \N^*$. The equation $z^n = a$ has exactly $n$ solutions:

$$
z_k = r^{1/n}\, \eu^{\iu\left(\frac{\alpha}{n} +
\frac{2k\pi}{n}\right)},
\qquad k = 0, 1, \dots, n - 1 .
$$

**Proof.** Write $z = \rho\,\eu^{\iu\theta}$ ($\rho > 0$). Then $z^n =
\rho^n \eu^{\iu n\theta} = r \eu^{\iu\alpha}$ if and only if $\rho^n
= r$ (moduli) and $n\theta \equiv \alpha \pmod{2\pi}$ (arguments), i.e. $\rho = r^{1/n}$ and $\theta = \frac{\alpha}{n} +
\frac{2k\pi}{n}$ for some $k \in \Z$. It remains to see when two integers $k, k'$ yield the same number: this happens exactly when the angles differ by a multiple of $2\pi$,

$$
\frac{2k\pi}n - \frac{2k'\pi}n \in 2\pi\Z
\iff \frac{k - k'}n \in \Z
\iff n \mid k - k' .
$$

By Euclidean division, every $k \in \Z$ is congruent mod $n$ to exactly one element of $\{0, 1, \dots, n-1\}$, so this range lists each solution once and the count is exactly $n$. (The same argument, run inside $\mathbb U_n$, shows the roots form a regular $n$-gon: consecutive values of $k$ rotate by the fixed angle $\frac{2\pi}n$.) ∎

**Example 3.15 (Cube roots of −27-27−27).**

Solve $z^3 = -27$. Exponential form of the right side: $-27 =
27\,\eu^{\iu\pi}$, so the three roots are

$$
z_k = 3\,\eu^{\iu(\frac\pi3 + \frac{2k\pi}3)}, \quad k = 0, 1, 2 :
\qquad
z_0 = 3\eu^{\iu\pi/3} = \frac32 + \frac{3\sqrt3}2\,\iu,
\quad
z_1 = -3,
\quad
z_2 = \conj{z_0} .
$$

Two checks. First, the real root $-3$ is the obvious one, and the other two are its rotations by $\pm\frac{2\pi}3$ — equivalently $-3j$ and $-3j^2$. Second, algebra confirms: $z^3 + 27 = (z +
3)(z^2 - 3z + 9)$, and the quadratic has discriminant $9 - 36 =
-27 < 0$ with roots $\frac{3 \pm 3\iu\sqrt3}2 = z_0, \conj{z_0}$. The insight: for real right-hand sides, the non-real roots always come in [conjugate](#def-b1-complex-field) pairs, so a picture of the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) is symmetric about the real axis — a preview of the real factorization theorem of [Chapter 8](https://one-course.com/books/math/3/en/chapter/8-polynomials#ch-b1-poly).

**Example 3.16 (A non-real right-hand side).**

Solve $z^4 = -8 + 8\iu\sqrt3$. Exponential form of the right side: [modulus](#def-b1-complex-field) $\sqrt{64 + 192} = 16$, argument $\theta$ with $\cos\theta = -\frac12$, $\sin\theta = \frac{\sqrt3}2$, i.e. $\theta = \frac{2\pi}3$. The four roots are

$$
z_k = 2\,\eu^{\iu(\frac\pi6 + \frac{k\pi}2)}, \quad k = 0, 1, 2,
3 :
\qquad
z_0 = \sqrt3 + \iu,\quad z_1 = \iu z_0 = -1 + \iu\sqrt3,
$$

$$
z_2 = -z_0 = -\sqrt3 - \iu,\qquad z_3 = -\iu z_0 = 1 - \iu\sqrt3 .
$$

(Check: $z_0^2 = 2 + 2\iu\sqrt3$, so $z_0^4 = (2 +
2\iu\sqrt3)^2 = 4 - 12 + 8\iu\sqrt3 = -8 + 8\iu\sqrt3$.) Once *one* root is found, the other three come free: they are its successive rotations by $\frac\pi2$, i.e. its products with the fourth [roots of unity](#def-b1-complex-unity) — the general structure behind [Theorem 3.14](#thm-b1-complex-roots), worth exploiting before recomputing each root from scratch. No [conjugate](#def-b1-complex-field) symmetry this time: the right-hand side is not real.

**Definition 3.17 (Roots of unity).**

The *$n$-th roots of unity* are the solutions of $z^n = 1$:

$$
\mathbb{U}_n = \bigl\{\, \omega^k : k = 0, \dots, n-1 \,\bigr\},
\qquad \omega = \eu^{2\iu\pi/n}.
$$

They form a group under multiplication ([Chapter 7](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#ch-b1-structures)) and sit at the vertices of a regular $n$-gon inscribed in the unit circle.

![The fifth roots of unity, = 2 π/5: a regular pentagon on the unit circle.](https://one-course.com/images/onecourse/chapters/math-3/b1-complex/fig-8a472e96bb29.svg)

*The fifth [roots of unity](#def-b1-complex-unity), $\omega = \eu^{2\iu\pi/5}$: a regular pentagon on the unit circle.*

**Proposition 3.18 (Sum of the roots of unity).**

For $n \geq 2$, the $n$-th [roots of unity](#def-b1-complex-unity) sum to zero: $\sum_{k=0}^{n-1} \omega^k = 0$.

**Proof.** Geometric sum with ratio $\omega \neq 1$: $\sum_{k=0}^{n-1} \omega^k = \frac{\omega^n - 1}{\omega - 1} = 0$ since $\omega^n = 1$. ∎

**Example 3.19 (Reading real and imaginary parts).**

Splitting $\sum_{k=0}^{n-1}\omega^k = 0$ into real and imaginary parts yields two trigonometric identities for free:

$$
\sum_{k=0}^{n-1}\cos\frac{2k\pi}n = 0,
\qquad
\sum_{k=0}^{n-1}\sin\frac{2k\pi}n = 0
\qquad (n \geq 2).
$$

Geometrically: the centroid of a regular $n$-gon inscribed in the unit circle is its center — the vertices balance out exactly. For $n = 5$ the first identity gives $1 + 2\cos\frac{2\pi}5 +
2\cos\frac{4\pi}5 = 0$ (pairing $k$ with $n - k$), the starting point of [Exercise 3.8](#exo-b1-complex-8)’s computation of $\cos\frac{2\pi}5$.

**Example 3.20 (Square roots in algebraic form).**

To solve $z^2 = 3 + 4\iu$ without trigonometry, set $z = x + \iu y$:

$$
x^2 - y^2 = 3, \qquad 2xy = 4, \qquad x^2 + y^2 = \abs{3 + 4\iu} = 5 .
$$

Adding the first and last: $x^2 = 4$, so $x = \pm 2$, then $y = 2/x =
\pm 1$ with the *same* sign pairing ($xy = 2 > 0$): $z = \pm(2 +
\iu)$. Combined with the usual formula, this solves every quadratic equation with complex coefficients ([Exercise 3.7](#exo-b1-complex-7)).

## 3.4 Complex numbers and plane geometry

**Proposition 3.21 (Geometric dictionary).**

Identify the point $M(x, y)$ of the plane with its *affix* $z = x
+ \iu y$. For distinct points $A, B, C$ of affixes $a, b,
c$:

1. $\abs{b - a}$ is the distance $AB$ ;
2. $\arg\dfrac{c - a}{b - a}$ is the angle between the vectors $\vect{AB}$ and $\vect{AC}$ (mod $2\pi$ );
3. $A, B, C$ are aligned if and only if $\dfrac{c - a}{b - a} \in  \R$ ; the lines $AB$ and $AC$ are perpendicular if and only if $\dfrac{c - a}{b - a} \in \iu\R$ .

**Proof.** (1) is the definition of the [modulus](#def-b1-complex-field) applied to $b - a$, the affix of $\vect{AB}$. (2): write $b - a = r\eu^{\iu\theta}$, $c - a =
s\eu^{\iu\varphi}$; then $\frac{c-a}{b-a} = \frac sr
\eu^{\iu(\varphi - \theta)}$ has argument $\varphi - \theta$, the angle from $\vect{AB}$ to $\vect{AC}$. (3): alignment means angle $0$ or $\pi$, i.e. argument in $\pi\Z$, i.e. the quotient is real; perpendicularity means angle $\pm\frac\pi2$, i.e. the quotient is purely imaginary. (The quotient is nonzero since $C \neq A$.) ∎

**Remark 3.22 (Interlude: C\CC is the plane with a multiplication).**

It is worth pausing on what makes this chapter possible at all: the plane $\R^2$ carries additions in every direction, but no God-given multiplication — and $\C$ *is* the plane equipped with one, in which multiplying by a fixed number rotates and scales. This one structure will be milked three more times in this volume. In [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices), multiplication by $a + \iu b$ reappears as the $2 \times 2$ matrix with rows $(a,
-b)$ and $(b, a)$: complex arithmetic is a first, entirely concrete family of matrix products. In [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid), the formula $\Re(\conj z\,w)$ turns out to be the dot product, and $\abs z$ the Euclidean norm: the triangle inequality proved here is the model for the Cauchy–Schwarz story there. And in [Chapter 24](https://one-course.com/books/math/3/en/chapter/24-plane-curves#ch-b1-curves), a moving point is best written $t \mapsto
z(t)$, so that velocity and acceleration become complex-valued derivatives — circular motion, for one, is simply $z(t) =
R\,\eu^{\iu\omega t}$. One good multiplication, four chapters of dividends.

**Proposition 3.23 (Maps z↦az+bz \mapsto az + bz↦az+b).**

Let $a \in \C^*$, $b \in \C$. The transformation $f(z) = az + b$ of the plane:

- is a translation if $a = 1$ ;
- otherwise has a unique fixed point $\zeta = \frac{b}{1-a}$ , and $f(z) - \zeta = a\,(z - \zeta)$ : $f$ is the rotation of center $\zeta$ and angle $\arg a$ , composed with the scaling of center $\zeta$ and ratio $\abs a$ .

In particular $z \mapsto \eu^{\iu\theta} z$ is the rotation of angle $\theta$ about the origin, and $\conj z$ is the reflection in the real axis.

**Proof.** If $a = 1$, $f(z) = z + b$ translates by the vector of affix $b$. If $a \neq 1$, the fixed-point equation $z = az + b$ has the unique solution $\zeta = \frac{b}{1-a}$, and then $f(z) - \zeta = az + b -
(a\zeta + b) = a(z - \zeta)$. Writing $a = \abs a\, \eu^{\iu\arg a}$, multiplication by $a$ scales distances to $\zeta$ by $\abs a$ and adds $\arg a$ to angles at $\zeta$, which is the announced composition. ∎

**Example 3.24 (Classifying a map z↦az+bz \mapsto az+bz↦az+b).**

Take $f(z) = \iu z + 1$. Here $a = \iu \neq 1$: the fixed point is

$$
\zeta = \frac{b}{1 - a} = \frac1{1 - \iu}
= \frac{1 + \iu}{2},
$$

and since $\abs a = 1$, there is no scaling at all: $f$ is the pure rotation of center $\frac{1+\iu}2$ and angle $\arg \iu = \frac\pi2$. Check: $f(\zeta) = \iu\,\frac{1+\iu}2 + 1 = \frac{\iu - 1}2 + 1 =
\frac{1 + \iu}2 = \zeta$, and $f(0) = 1$, $f(1) = 1 + \iu$, $f(1 +
\iu) = \iu\,(1 + \iu) + 1 = \iu$: the four points $0, 1, 1+\iu, \iu$ of the unit square cycle around their center $\zeta$, one quarter turn at a time — exactly what a rotation by $\frac\pi2$ about the square’s center must do.

![The map f(z) = z + 1 of : a quarter-turn rotation about the fixed point = 1+ 2. The vertices of the unit square cycle 0 1 1+ 0; no point moves along a straight line, yet the whole square is rigidly turned.](https://one-course.com/images/onecourse/chapters/math-3/b1-complex/fig-93c487327b70.svg)

*The [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $f(z) = \iu z + 1$ of [Example 3.24](#ex-b1-complex-classify): a quarter-turn rotation about the fixed point $\zeta = \frac{1+\iu}2$. The vertices of the unit square cycle $0 \to 1 \to 1{+}\iu \to \iu \to 0$; no point moves along a straight line, yet the whole square is rigidly turned.*

**Remark 3.25 (Common pitfalls with moduli and arguments).**

1. *No inequalities in $\C$.* Writing $z \leq w$ for non-real numbers is meaningless; only moduli, real parts and imaginary parts can be compared.
2. *$\sqrt{\phantom z}$ is reserved for $\R_+$.* Every nonzero complex number has *two* square roots and neither is privileged: write “let $\delta$ be a square root of $\Delta$ ” (computed as in [Example 3.20](#ex-b1-complex-sqrt) ), never $\sqrt\Delta$ — the rule $\sqrt{ab} = \sqrt a\sqrt b$ already fails at $a = b = -1$ .
3. *Arguments live modulo $2\pi$.* From $\eu^{\iu\alpha} = \eu^{\iu\beta}$ conclude $\alpha  \equiv \beta \pmod{2\pi}$ , not $\alpha = \beta$ ; forgetting the $2k\pi$ is how solution [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $z^n = a$ lose $n - 1$ of their $n$ elements.
4. *$\abs{z + w}$ is not $\abs z + \abs w$.* Equality in the triangle inequality is the exceptional aligned case ( [Proposition 3.2](#prop-b1-complex-rules) (4)); in general the [modulus](#def-b1-complex-field) of a sum must be estimated, not computed.

**Example 3.26.**

$A, B, C$ (affixes $a, b, c$, pairwise distinct) form an equilateral triangle with the vertices in direct (counterclockwise) order if and only if $\frac{c - a}{b - a} = \eu^{\iu\pi/3}$: the rotation of center $A$ and angle $\frac\pi3$ sends $B$ to $C$. Both orientations together are captured by the symmetric equation $a^2 + b^2 + c^2 = ab
+ bc + ca$ ([Exercise 3.10](#exo-b1-complex-10)).

**Remark 3.27 (Where this chapter is used).**

The exponential form is the single most reused computational device of the volume. [Roots of unity](#def-b1-complex-unity) become the standard example of a cyclic group in [Chapter 7](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#ch-b1-structures) and drive the factorization of $X^n - 1$ in [Chapter 8](https://one-course.com/books/math/3/en/chapter/8-polynomials#ch-b1-poly); grouping [conjugate](#def-b1-complex-field) roots there produces the real factorizations used by partial fractions in [Chapter 9](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#ch-b1-fractions). Linearization ([Method 3.11](#met-b1-complex-trig)) is the standard preparation for integrating trigonometric powers in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration), and the characteristic equations of [Chapter 5](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#ch-b1-diffeq) have complex roots whose real and imaginary parts produce the oscillating solutions $\eu^{\lambda t}\cos\omega
t$. The geometric dictionary returns in matrix clothing: rotations and similarities become the orthogonal matrices of Chapters [21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) and [23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid), and parametrized curves in [Chapter 24](https://one-course.com/books/math/3/en/chapter/24-plane-curves#ch-b1-curves) are often best written as [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $t \mapsto z(t)
\in \C$.

## 3.5 Exercises

**Exercise 3.1 ★.**

Put in exponential form: $1 + \iu$; $\;\sqrt 3 - \iu$; $\;-5$; $\;\dfrac{1 + \iu}{\sqrt 3 - \iu}$. Deduce $\cos\frac{5\pi}{12}$ and $\sin\frac{5\pi}{12}$.

**Solution of Exercise 3.1.**

$1 + \iu = \sqrt 2\, \eu^{\iu\pi/4}$; $\;\sqrt 3 - \iu = 2\, \eu^{-\iu\pi/6}$; $\;-5 = 5\, \eu^{\iu\pi}$;

$$
\frac{1 + \iu}{\sqrt 3 - \iu}
= \frac{\sqrt 2}{2}\, \eu^{\iu(\pi/4 + \pi/6)}
= \frac{\sqrt 2}{2}\, \eu^{5\iu\pi/12}.
$$

Computing the same quotient algebraically (multiply by the [conjugate](#def-b1-complex-field)):

$$
\frac{(1 + \iu)(\sqrt 3 + \iu)}{4}
= \frac{(\sqrt 3 - 1) + \iu(\sqrt 3 + 1)}{4}.
$$

Identifying with $\frac{\sqrt 2}{2}(\cos\frac{5\pi}{12} +
\iu\sin\frac{5\pi}{12})$:

$$
\cos\frac{5\pi}{12} = \frac{\sqrt 6 - \sqrt 2}{4},
\qquad
\sin\frac{5\pi}{12} = \frac{\sqrt 6 + \sqrt 2}{4}.
$$

**Exercise 3.2 ★.**

Compute $(1 + \iu)^{20}$. For which $n \in \N$ is $(1 + \iu)^n$ a real number?

**Solution of Exercise 3.2.**

$(1+\iu)^2 = 2\iu$, so $(1+\iu)^{20} = (2\iu)^{10} = 2^{10}\,\iu^{10}
= 1024 \times (-1) = -1024$.

In exponential form $(1+\iu)^n = 2^{n/2}\, \eu^{\iu n\pi/4}$, real if and only if $\sin\frac{n\pi}{4} = 0$, i.e. $4 \mid n$. So $(1+\iu)^n
\in \R$ exactly for the multiples of $4$ (value $(-4)^{n/4}$).

**Exercise 3.3 ★.**

Describe geometrically the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $z \in \C$ such that: $\abs{z - 2} = \abs{z + \iu}$; $\;\abs{z - 1} = 2$; $\;\dfrac{z - 1}{z + 1} \in \iu\R$ (for $z \neq -1$).

**Solution of Exercise 3.3.**

$\abs{z - 2} = \abs{z + \iu}$: equidistance from the points $2$ and $-\iu$ — the perpendicular bisector of the segment joining $(2, 0)$ and $(0, -1)$.

$\abs{z - 1} = 2$: the circle of center $1$ and radius $2$.

$\frac{z-1}{z+1} \in \iu\R$: by [Proposition 3.21](#prop-b1-complex-geometry) (3), the points $M(z)$, $A(1)$, $B(-1)$ satisfy: lines $MA$ and $MB$ perpendicular (or $z = 1$, where the quotient is $0 \in \iu\R$). The [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) is the circle of diameter $[-1, 1]$ (the unit circle), minus the point $-1$ where the quotient is undefined. *Check by computation:* $z = \eu^{\iu\theta}$ gives $\frac{z-1}{z+1} = \frac{\eu^{\iu\theta/2}(\eu^{\iu\theta/2} -
\eu^{-\iu\theta/2})}{\eu^{\iu\theta/2}(\eu^{\iu\theta/2} +
\eu^{-\iu\theta/2})} = \iu\tan\frac\theta2 \in \iu\R$.

**Exercise 3.4 ★.**

Linearize $\sin^4\theta$, and expand $\cos 4\theta$ as a polynomial in $\cos\theta$.

**Solution of Exercise 3.4.**

Linearization:

$$
\sin^4\theta
= \Bigl(\frac{\eu^{\iu\theta} - \eu^{-\iu\theta}}{2\iu}\Bigr)^{\!4}
= \frac{\eu^{4\iu\theta} - 4\eu^{2\iu\theta} + 6 - 4\eu^{-2\iu\theta}
+ \eu^{-4\iu\theta}}{16}
= \frac{\cos 4\theta - 4\cos 2\theta + 3}{8}.
$$

Expansion: by de Moivre, $\cos 4\theta = \Re\bigl((c +
\iu s)^4\bigr) = c^4 - 6c^2 s^2 + s^4$ with $c = \cos\theta$, $s =
\sin\theta$; substituting $s^2 = 1 - c^2$:

$$
\cos 4\theta = c^4 - 6c^2(1 - c^2) + (1 - c^2)^2
= 8c^4 - 8c^2 + 1 .
$$

**Exercise 3.5 ★.**

Solve $z^3 = 8\iu$ and place the solutions on a picture. Solve $z^4 = -4$ and factor $X^4 + 4$ into two real quadratic polynomials.

**Solution of Exercise 3.5.**

$8\iu = 8\,\eu^{\iu\pi/2}$, so the cube roots are $2\,\eu^{\iu(\pi/6 +
2k\pi/3)}$, $k = 0, 1, 2$:

$$
z_0 = 2\eu^{\iu\pi/6} = \sqrt 3 + \iu,\quad
z_1 = 2\eu^{5\iu\pi/6} = -\sqrt 3 + \iu,\quad
z_2 = 2\eu^{3\iu\pi/2} = -2\iu :
$$

an equilateral triangle on the circle of radius $2$.

$-4 = 4\eu^{\iu\pi}$, so $z^4 = -4$ has solutions $\sqrt 2\,
\eu^{\iu(\pi/4 + k\pi/2)}$: $\;1 + \iu$, $-1 + \iu$, $-1 - \iu$, $1 - \iu$. Pairing [conjugate](#def-b1-complex-field) roots:

$$
X^4 + 4 = \bigl(X^2 - 2X + 2\bigr)\bigl(X^2 + 2X + 2\bigr),
$$

since $(X - (1+\iu))(X - (1-\iu)) = X^2 - 2X + 2$ and similarly for the other pair. (Expand to check.)

**Exercise 3.6 ★★.**

For $\theta \in \R$ with $\eu^{\iu\theta} \neq 1$ and $n \in \N$, compute

$$
C_n = \sum_{k=0}^{n} \cos k\theta
\qquad\text{and}\qquad
S_n = \sum_{k=0}^{n} \sin k\theta
$$

by summing the geometric series $\sum_k \eu^{\iu k\theta}$ and using the half-angle factorization.

**Solution of Exercise 3.6.**

$C_n + \iu S_n = \sum_{k=0}^{n} \eu^{\iu k\theta} =
\frac{\eu^{\iu(n+1)\theta} - 1}{\eu^{\iu\theta} - 1}$ (geometric sum, ratio $\eu^{\iu\theta} \neq 1$). Half-angle factorization ([Method 3.11](#met-b1-complex-trig) (4)) on numerator and denominator:

$$
\frac{2\iu\sin\frac{(n+1)\theta}{2}\, \eu^{\iu(n+1)\theta/2}}
{2\iu\sin\frac{\theta}{2}\, \eu^{\iu\theta/2}}
= \frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\,
\eu^{\iu n\theta/2}.
$$

Taking real and imaginary parts:

$$
C_n = \frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}
\cos\frac{n\theta}{2},
\qquad
S_n = \frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}
\sin\frac{n\theta}{2}.
$$

**Exercise 3.7 ★★.**

Solve in $\C$: $z^2 - (3 + 4\iu) z + (-1 + 5\iu) = 0$. *(Compute the discriminant, extract its square roots as in [Example 3.20](#ex-b1-complex-sqrt).)*

**Solution of Exercise 3.7.**

Discriminant: $\Delta = (3 + 4\iu)^2 - 4(-1 + 5\iu) = 9 + 24\iu - 16 +
4 - 20\iu = -3 + 4\iu$. Square roots of $-3 + 4\iu$: solve $x^2 - y^2
= -3$, $2xy = 4$, $x^2 + y^2 = 5$; then $x^2 = 1$, $y = 2/x$, same signs: $\delta = \pm(1 + 2\iu)$. Hence

$$
z = \frac{(3 + 4\iu) \pm (1 + 2\iu)}{2}
\in \{\, 2 + 3\iu,\; 1 + \iu \,\}.
$$

*Check:* sum $= 3 + 4\iu$ and product $(2+3\iu)(1+\iu) = -1 +
5\iu$, as required by the coefficients.

**Exercise 3.8 ★★.**

Let $\omega = \eu^{2\iu\pi/5}$.

1. Justify $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ .
2. Set $u = \omega + \omega^4$ and $v = \omega^2 + \omega^3$ . Compute $u + v$ and $uv$ , and deduce that $u$ and $v$ are the roots of $X^2 + X - 1$ .
3. Conclude that $\cos\frac{2\pi}{5} = \frac{\sqrt 5 - 1}{4}$ .

**Solution of Exercise 3.8.**

1. [Proposition 3.18](#prop-b1-complex-sumroots) with $n = 5$ .
2. $u + v = \omega + \omega^2 + \omega^3 + \omega^4 = -1$ by (1). For the product, expand and reduce exponents modulo $5$: $$uv = (\omega + \omega^4)(\omega^2 + \omega^3)  = \omega^3 + \omega^4 + \omega^6 + \omega^7  = \omega^3 + \omega^4 + \omega + \omega^2 = -1 .$$ So $u$ and $v$ have sum $-1$ and product $-1$: they are the two roots of $X^2 + X - 1$.
3. $u = \omega + \conj\omega = 2\cos\frac{2\pi}{5} > 0$ (the angle is acute), and the positive root of $X^2 + X - 1$ is $\frac{-1 + \sqrt 5}{2}$ . Hence $\cos\frac{2\pi}{5} =  \frac{\sqrt 5 - 1}{4}$ .

**Exercise 3.9 ★★.**

Prove that for all $z, w \in \C$ (parallelogram identity):

$$
\abs{z + w}^2 + \abs{z - w}^2 = 2\abs z^2 + 2\abs w^2 ,
$$

and interpret it geometrically in the parallelogram with vertices $0,
z, w, z + w$.

**Solution of Exercise 3.9.**

Expand both squares as in the proof of [Proposition 3.2](#prop-b1-complex-rules) (4):

$$
\abs{z + w}^2 = \abs z^2 + \abs w^2 + 2\Re(z\conj w),
\qquad
\abs{z - w}^2 = \abs z^2 + \abs w^2 - 2\Re(z\conj w),
$$

and add. Geometrically, $\abs{z+w}$ and $\abs{z-w}$ are the lengths of the two diagonals of the parallelogram with vertices $0, z, z+w, w$, while $\abs z$ and $\abs w$ are the side lengths: the sum of the squares of the diagonals equals the sum of the squares of the four sides.

**Exercise 3.10 ★★★.**

Prove that three pairwise distinct points of affixes $a, b, c$ form an equilateral triangle (either orientation) if and only if

$$
a^2 + b^2 + c^2 = ab + bc + ca .
$$

*Hint: the direct case is $\frac{c-a}{b-a} = -j^2$ and the indirect case $\frac{c-a}{b-a} = -j$, where $j = \eu^{2\iu\pi/3}$ satisfies $j^2 + j + 1 = 0$; or factor $a + jb + j^2c$ and $a + j^2 b
+ jc$.*

**Solution of Exercise 3.10.**

Let $j = \eu^{2\iu\pi/3}$, so $j^2 + j + 1 = 0$ and $\eu^{\iu\pi/3} =
-j^2$, $\eu^{-\iu\pi/3} = -j$. The triangle is equilateral direct iff $c - a = -j^2 (b - a)$, indirect iff $c - a = -j(b - a)$ ([Example 3.26](#ex-b1-complex-equilateral)).

Consider $P = a + jb + j^2 c$ and $Q = a + j^2 b + jc$. Using $1 +
j^2 = -j$ and $j^3 = 1$:

$$
c - a + j^2(b - a) = c + j^2 b - (1 + j^2)\,a = c + j^2 b + ja
= j\,(a + jb + j^2 c) = jP,
$$

so the direct case reads $jP = 0 \iff P = 0$; the same computation with $j$ in place of $j^2$ gives $c - a + j(b - a) = j^2 Q$, so the indirect case reads $Q = 0$. Hence: equilateral (either orientation) $\iff PQ = 0$. Expanding, with $j + j^2 = -1$:

$$
PQ = a^2 + b^2 + c^2 + (j + j^2)(ab + bc + ca)
= a^2 + b^2 + c^2 - (ab + bc + ca).
$$

So the triangle is equilateral if and only if $a^2 + b^2 + c^2 = ab +
bc + ca$.

**Exercise 3.11 ★★★.**

For $n \in \N^*$, evaluate $P = \prod_{k=1}^{n-1}
\bigl(1 - \omega^k\bigr)$ where $\omega = \eu^{2\iu\pi/n}$. *Hint: $X^n - 1 = \prod_{k=0}^{n-1} (X - \omega^k)$; divide by $X - 1$ and evaluate at $X = 1$.* Deduce $\prod_{k=1}^{n-1} \sin\frac{k\pi}{n} = \dfrac{n}{2^{n-1}}$.

**Solution of Exercise 3.11.**

Since the $\omega^k$, $k = 0, \dots, n-1$, are exactly the $n$ roots of $X^n - 1$ ([Theorem 3.14](#thm-b1-complex-roots)), and the polynomial is monic:

$$
X^n - 1 = \prod_{k=0}^{n-1} (X - \omega^k)
= (X - 1) \prod_{k=1}^{n-1} (X - \omega^k).
$$

Dividing by $X - 1$: $\;1 + X + \dots + X^{n-1} = \prod_{k=1}^{n-1}
(X - \omega^k)$. Evaluating at $X = 1$ gives $P = n$.

Now $1 - \omega^k = -\eu^{\iu k\pi/n}\bigl(\eu^{\iu k\pi/n} -
\eu^{-\iu k\pi/n}\bigr) = -2\iu\,\eu^{\iu k\pi/n} \sin\frac{k\pi}{n}$, so taking moduli in $P = n$ (each $\sin\frac{k\pi}n > 0$ for $1 \leq k
\leq n-1$):

$$
n = \abs P = \prod_{k=1}^{n-1} 2\sin\frac{k\pi}{n}
= 2^{n-1} \prod_{k=1}^{n-1} \sin\frac{k\pi}{n},
\qquad\text{hence}\qquad
\prod_{k=1}^{n-1} \sin\frac{k\pi}{n} = \frac{n}{2^{n-1}} .
$$

**Exercise 3.12 ★★.**

Let $n \geq 2$. Solve the equation $(z + 1)^n = (z - 1)^n$ in $\C$: show that it has exactly $n - 1$ solutions, all purely imaginary, namely

$$
z_k = -\iu\,\frac{\cos\frac{k\pi}{n}}{\sin\frac{k\pi}{n}},
\qquad k = 1, \dots, n - 1 .
$$

*Hint: $z = 1$ is not a solution, so divide and use the [roots of unity](#def-b1-complex-unity); then apply the half-angle factorization of [Method 3.11](#met-b1-complex-trig).*

**Solution of Exercise 3.12.**

$z = 1$ is not a solution ($2^n \neq 0$), so the equation is equivalent to $\bigl(\frac{z+1}{z-1}\bigr)^n = 1$, i.e. $\frac{z+1}{z-1} = \omega^k$ with $\omega = \eu^{2\iu\pi/n}$ and $k \in \{0, \dots, n-1\}$. The value $k = 0$ is excluded ($z + 1 =
z - 1$ is impossible). For $1 \leq k \leq n - 1$, solving $z + 1 = \omega^k(z - 1)$ gives $z(1 - \omega^k) = -1 - \omega^k$, so with $\varphi = \frac{2k\pi}{n}$ and the half-angle factorizations $1 + \eu^{\iu\varphi} = 2\cos\frac\varphi2\,
\eu^{\iu\varphi/2}$, $\eu^{\iu\varphi} - 1 = 2\iu\sin\frac\varphi2
\,\eu^{\iu\varphi/2}$:

$$
z_k = \frac{1 + \omega^k}{\omega^k - 1}
= \frac{2\cos\frac{k\pi}{n}}{2\iu\,\sin\frac{k\pi}{n}}
= -\iu\,\frac{\cos\frac{k\pi}{n}}{\sin\frac{k\pi}{n}} ,
$$

purely imaginary as claimed. The [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $t \mapsto \cos t/\sin t$ is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) on $\intoo0\pi$ (it is strictly decreasing), so the $n - 1$ values $z_k$ are pairwise distinct: the equation, of degree $n - 1$ once expanded (the $z^n$ terms cancel), has exactly these $n - 1$ solutions.

## 3.6 Problem: Napoleon’s theorem

**Problem 3.1.**

Erect an equilateral triangle outward on each side of an *arbitrary* triangle: the three centers of those triangles always form an equilateral triangle. This is Napoleon’s theorem — a [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) with no visible reason to be true, which the algebra of $j = \eu^{2\iu\pi/3}$ proves in three lines of computation. This problem builds the full toolkit (rotations, direct similarities, the $j$-criterion of [Exercise 3.10](#exo-b1-complex-10)), proves Napoleon’s theorem inner and outer, locates the degenerate case, and closes with a second jewel of the same school: van Schooten’s theorem on equilateral triangles and Ptolemy’s inequality. Throughout, points of the plane are identified with their affixes.

**Part I — The number $j$ and rotations.**

1. Compute the algebraic form of $j$ , then $j^2$ , $j^3$ , $1 + j  + j^2$ , $\conj j$ and $j^{-1}$ . Place $1$ , $j$ , $j^2$ on a sketch of the unit circle.
2. Show that the rotation of center $a$ and angle $\theta$ is $r(z) = a + \eu^{\iu\theta}(z - a)$ , and that the composition of two rotations, of angles $\theta$ and $\theta'$ , is a rotation of angle $\theta + \theta'$ if $\theta + \theta' \notin 2\pi\Z$ , and a translation otherwise. (Use [Proposition 3.23](#prop-b1-complex-similitude) .)
3. Let $b \neq c$ . Show that there are exactly two points $p$ making $(b, c, p)$ equilateral, namely $p = b +  \eu^{\pm\iu\pi/3}(c - b)$ . For a *direct* triangle $(a, b, c)$ , check on the example $a = 0$ , $b = 1$ , $c =  \iu$ that the choice $\eu^{-\iu\pi/3}$ is the one lying on the far side of the line $BC$ from $a$ — the *outward* apex.
4. From the solution of [Exercise 3.10](#exo-b1-complex-10), $(a, b, c)$ is direct equilateral iff $a + jb + j^2c = 0$. Prove the two rotation identities $$b + jc + j^2a = j^2\,(a + jb + j^2c),  \qquad  c + ja + j^2b = j\,(a + jb + j^2c),$$ and deduce that the criterion is invariant under cyclic [permutation](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-objects) of $(a, b, c)$.
5. Show that if $a + jb + j^2c = 0$ and two of the three points coincide, then all three coincide. (So the criterion exactly characterizes: direct equilateral triangle, or a single point.)

**Part II — Direct similarities.**

6. Show that the [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $f(z) = az + b$ with $a \in \C^*$ ( *direct similarities* ) are stable under composition and inversion: the composition of two of them, and the inverse of any of them, is again of this form. (In the language of [Chapter 7](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#ch-b1-structures) : they form a group.)
7. Given $z_1 \neq z_2$ and $w_1 \neq w_2$ , show there is a *unique* direct similarity $f$ with $f(z_1) = w_1$ and $f(z_2) = w_2$ , and give $a$ and $b$ explicitly.
8. Show that a direct similarity $f(z) = az + b$ multiplies all distances by $\abs a$ and preserves the *shape* $\frac{c - a'}{b' - a'}$ of any triangle $(a', b', c')$ *(numerator and denominator are both multiplied by $a$)* . Deduce that two triangles are directly similar exactly when their shapes are equal.
9. Determine completely the direct similarity with $f(0) = 1$ and $f(1) = \iu$ : give $a$ , $b$ , the fixed point, the ratio and the angle.
10. Let $\alpha + \beta = 1$ . Show that the “weighted point” $g(a', b') = \alpha a' + \beta b'$ commutes with every direct similarity: $f(\alpha a' + \beta b') = \alpha f(a')  + \beta f(b')$ . Deduce that centroids, midpoints, and the Napoleon centers below are all transported by similarities — the algebraic license behind every “without loss of generality, place the circumcircle at the unit circle” argument.

**Part III — Napoleon’s theorem.** Let $(a, b, c)$ be a direct triangle. On each side erect the outward equilateral triangle (question 3) and let $n_a$, $n_b$, $n_c$ be the centers (centroids) of the triangles erected on $[b, c]$, $[c, a]$, $[a, b]$ respectively.

11. Show that $$n_a = \alpha b + \beta c, \qquad  n_b = \alpha c + \beta a, \qquad  n_c = \alpha a + \beta b,  \qquad\text{with}\quad  \alpha = \frac{3 + \iu\sqrt3}{6},\ \beta = \conj\alpha .$$
12. Check $\alpha + \beta = 1$ , and deduce that the triangle $(n_a, n_b, n_c)$ has the *same centroid* as $(a, b, c)$ .
13. Using the identities of question 4, show that $$n_a + j\,n_b + j^2 n_c  = (\alpha j^2 + \beta j)\,(a + jb + j^2c) .$$
14. Compute $\alpha j + \beta$ and conclude: $\alpha j^2 +  \beta j = j(\alpha j + \beta) = 0$ , so $n_a + j n_b + j^2  n_c = 0$ for *every* triangle: the outer Napoleon triangle is direct equilateral (or a point). Napoleon’s theorem is proved.
15. Erect the equilateral triangles *inward* instead (choice $\eu^{+\iu\pi/3}$ in question 3) and let $m_a, m_b,  m_c$ be their centers. Show $m_a = \beta b + \alpha c$ (and cyclically), then prove $m_a + j^2 m_b + j\,m_c = 0$ : the inner Napoleon triangle is equilateral too, with the opposite orientation.
16. Locate the degeneracy: show that $n_a = n_b = n_c$ happens exactly when $a + j^2 b + jc = 0$ , i.e. when $(a, b, c)$ is an *indirect* equilateral triangle. *(Two points among $n_a, n_b, n_c$ coincide iff all three do, by question 5; then use $n_a + j^2 n_b + j n_c = j(\alpha +  \beta j)(a + j^2b + jc)$ and check $\alpha + \beta j \neq  0$.)*
17. Carry out the whole computation on the triangle $a = 0$ , $b = 1$ , $c = \iu$ : give $n_a, n_b, n_c$ exactly, and verify by direct computation of the three squared side lengths that the triangle is equilateral, of squared side $\frac{2 + \sqrt3}{3}$ .

**Part IV — Ptolemy and van Schooten.**

18. Prove the identity, valid for all complex $a, b, c, d$: $$(a - b)(c - d) + (a - d)(b - c) = (a - c)(b - d) .$$
19. Deduce *Ptolemy’s inequality*: for any four points $A, B, C, D$, $$AC \cdot BD \;\leq\; AB \cdot CD + AD \cdot BC ,$$ with equality if and only if $(a-b)(c-d)$ and $(a-d)(b-c)$ lie on a common ray from $0$.
20. Show that for $\theta, \varphi \in \R$ , $\abs{\eu^{\iu\theta} - \eu^{\iu\varphi}} =  2\,\abs{\sin\frac{\theta - \varphi}2}$ .
21. (van Schooten’s theorem) Let $(a, b, c) = (1, j, j^2)$ — by Part II this loses no generality among direct equilateral triangles — and let $p = \eu^{\iu\theta}$ with $\theta \in  \intoo{2\pi/3}{4\pi/3}$, a point of the circumcircle on the arc $BC$ not containing $A$. Prove $$PA = PB + PC .$$ *(Express the three distances with question 20 and use a sum-to-product formula.)*
22. Verify at $p = -1$ that this is exactly the equality case of Ptolemy’s inequality for the cyclic order $A, B, P, C$ : compute $(a - b)(p - c)$ and $(a - c)(b - p)$ and check that their ratio is a positive real.

**Part V — Synthesis.**

23. Compute the outer Napoleon triangle of the equilateral triangle $(1, j, j^2)$ itself, and describe the result geometrically.
24. Where exactly did the problem use: (i) multiplication as rotation; (ii) the group structure of similarities and the shape invariant; (iii) the half-angle factorization of [Method 3.11](#met-b1-complex-trig) ? One sentence each.
25. In a short paragraph, articulate the moral of the problem: what does the dictionary between plane geometry and the algebra of $\C$ buy, what does it cost, and which of the two proofsteps — the identity $\alpha j + \beta = 0$ or the classical picture — *explains* Napoleon’s theorem better in your view? Mention one place where the dictionary will reappear in matrix form later in this volume.

**Solution of Problem 3.1.**

**1.** $j = \cos\frac{2\pi}3 + \iu\sin\frac{2\pi}3 = -\frac12
+ \iu\frac{\sqrt3}2$; $j^2 = \eu^{4\iu\pi/3} = -\frac12 -
\iu\frac{\sqrt3}2 = \conj j$; $j^3 = 1$; $1 + j + j^2 = 0$ (sum of the cube [roots of unity](#def-b1-complex-unity), [Proposition 3.18](#prop-b1-complex-sumroots)); $j^{-1}
= j^2$ (since $j \cdot j^2 = 1$). On the unit circle, $1$, $j$, $j^2$ are the vertices of a direct equilateral triangle.

**2.** The rotation of center $a$ and angle $\theta$ fixes $a$ and turns every vector issued from $a$ by $\theta$: $r(z) - a =
\eu^{\iu\theta}(z - a)$, i.e. $r(z) = a + \eu^{\iu\theta}(z - a)$. Composing $r(z) = a + \eu^{\iu\theta}(z-a)$ and $r'(z) = a' +
\eu^{\iu\theta'}(z-a')$:

$$
r' \circ r\,(z) = \eu^{\iu(\theta + \theta')} z +
\text{constant},
$$

a [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) of the form $Az + B$ with $A = \eu^{\iu(\theta+\theta')}$ of [modulus](#def-b1-complex-field) $1$. By [Proposition 3.23](#prop-b1-complex-similitude), it is a rotation of angle $\arg A = \theta + \theta'$ if $A \neq 1$, and a translation if $A = 1$, i.e. if $\theta + \theta' \in 2\pi\Z$.

**3.** $(b, c, p)$ is equilateral iff $\abs{p - b} = \abs{c -
b} = \abs{p - c}$. Writing $q = \frac{p - b}{c - b}$, the first equality says $\abs q = 1$ and the second $\abs{q - 1} = 1$; together, $q = \eu^{\pm\iu\pi/3}$ (the two intersection points of the circles $\abs q = 1$ and $\abs{q - 1} = 1$ are $\frac12 \pm
\iu\frac{\sqrt3}2$). Hence $p = b + \eu^{\pm\iu\pi/3}(c - b)$. For $a = 0$, $b = 1$, $c = \iu$ (a direct triangle): the choice $\eu^{-\iu\pi/3}$ gives

$$
p = 1 + \Bigl(\tfrac12 - \iu\tfrac{\sqrt3}2\Bigr)(\iu - 1)
= \tfrac{1 + \sqrt3}2\,(1 + \iu) \approx 1.37 + 1.37\,\iu ,
$$

which lies on the far side of the line $BC$ ($x + y = 1$) from $a
= 0$: outward. The choice $\eu^{+\iu\pi/3}$ gives $p \approx -0.37
- 0.37\,\iu$, on the same side as $a$: inward.

**4.** Multiply $P = a + jb + j^2c$ by $j^2$: $j^2 P = j^2 a +
j^3 b + j^4 c = b + jc + j^2 a$ (using $j^3 = 1$, $j^4 = j$); and by $j$: $jP = ja + j^2 b + c$. These are the two identities. If $P
= 0$ then $j^2 P = jP = 0$: the criterion holds for $(b, c, a)$ and $(c, a, b)$ as well — cyclic invariance (as it must be: an equilateral triangle does not care which vertex is listed first).

**5.** If $a = b$: $0 = a(1 + j) + j^2 c = -j^2 a + j^2 c$ (using $1 + j = -j^2$), so $c = a$. If $b = c$: $0 = a + b(j + j^2)
= a - b$, so $a = b$. If $a = c$: $0 = a(1 + j^2) + jb = -ja + jb$, so $a = b$. In each case all three points coincide.

**6.** $(a'z + b') \circ (az + b) = a'a\,z + (a'b + b')$ with $a'a \neq 0$: same form. The inverse of $z \mapsto az + b$ is $z
\mapsto \frac1a z - \frac ba$, again of the same form. With the identity [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) as neutral element, the direct similarities form a group under composition.

**7.** $f(z) = az + b$ satisfies $f(z_1) = w_1$, $f(z_2) =
w_2$ iff $a z_1 + b = w_1$ and $a z_2 + b = w_2$; subtracting, $a(z_2 - z_1) = w_2 - w_1$, so

$$
a = \frac{w_2 - w_1}{z_2 - z_1} \;(\neq 0), \qquad
b = w_1 - a z_1
$$

are forced, and conversely this choice works: existence and uniqueness.

**8.** $\abs{f(z) - f(w)} = \abs{a(z - w)} = \abs a\,\abs{z -
w}$: all distances are multiplied by $\abs a$. For the shape:

$$
\frac{f(c') - f(a')}{f(b') - f(a')} = \frac{a(c' -
a')}{a(b' - a')} = \frac{c' - a'}{b' - a'} .
$$

If two triangles $(a', b', c')$ and $(a'', b'', c'')$ have equal shapes, let $f$ be the unique direct similarity with $f(a') = a''$ and $f(b') = b''$ (question 7); then the shape of $(a'', b'',
f(c'))$ equals that of $(a', b', c')$, hence that of $(a'', b'',
c'')$, and the shape determines the third vertex from the first two: $f(c') = c''$. Conversely equal shapes follow from the displayed invariance.

**9.** $b = f(0) = 1$; $a + b = f(1) = \iu$ gives $a = \iu -
1$. Ratio $\abs a = \sqrt2$, angle $\arg(\iu - 1) =
\frac{3\pi}4$. Fixed point:

$$
\zeta = \frac{b}{1 - a} = \frac1{2 - \iu} = \frac{2 + \iu}5 .
$$

So $f$ is the direct similarity of center $\frac{2+\iu}5$, ratio $\sqrt2$ and angle $\frac{3\pi}4$.

**10.** With $\alpha + \beta = 1$ and $f(z) = az + b$:

$$
f(\alpha a' + \beta b') = a\alpha a' + a\beta b' + b
= \alpha(a a' + b) + \beta(a b' + b)
= \alpha f(a') + \beta f(b') ,
$$

the key being $b = (\alpha + \beta)b$. Midpoints ($\alpha = \beta =
\frac12$), centroids (iterate), and the Napoleon centers $\alpha b
+ \beta c$ below are therefore *equivariant*: transforming the triangle transforms them accordingly. Hence proving a similarity-invariant [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) for one well-placed triangle proves it for all — the license used in question 21.

**11.** The outward apex on $[b, c]$ is $p_a = b +
\eu^{-\iu\pi/3}(c - b)$ (question 3), so the center is

$$
n_a = \frac{b + c + p_a}3
= \frac{2b + c + \frac{1 - \iu\sqrt3}2\,(c - b)}3
= \frac{(3 + \iu\sqrt3)\,b + (3 - \iu\sqrt3)\,c}6
= \alpha b + \beta c ,
$$

with $\alpha = \frac{3 + \iu\sqrt3}6$ and $\beta = \frac{3 -
\iu\sqrt3}6 = \conj\alpha$. The same computation on the sides $[c, a]$ and $[a, b]$ gives $n_b = \alpha c + \beta a$ and $n_c =
\alpha a + \beta b$ (cyclic shift of the roles).

**12.** $\alpha + \beta = \frac{3 + \iu\sqrt3 + 3 -
\iu\sqrt3}6 = 1$. Hence

$$
\frac{n_a + n_b + n_c}3
= \frac{(\alpha + \beta)(a + b + c)}3 = \frac{a + b + c}3 :
$$

the two triangles share their centroid.

**13.** Group by $\alpha$ and $\beta$ and apply question 4’s identities to $P = a + jb + j^2c$:

$$
n_a + j n_b + j^2 n_c
= \alpha\,(b + jc + j^2 a) + \beta\,(c + ja + j^2 b)
= \alpha\,j^2 P + \beta\,j P
= (\alpha j^2 + \beta j)\,P .
$$

**14.** With $j = \frac{-1 + \iu\sqrt3}2$:

$$
\alpha j = \frac{(3 + \iu\sqrt3)(-1 + \iu\sqrt3)}{12}
= \frac{-3 + 3\iu\sqrt3 - \iu\sqrt3 - 3}{12}
= \frac{-3 + \iu\sqrt3}6 = -\beta ,
$$

so $\alpha j + \beta = 0$, hence $\alpha j^2 + \beta j = j(\alpha j
+ \beta) = 0$, and question 13 gives $n_a + jn_b + j^2n_c = 0$ for every triangle $(a, b, c)$. By the criterion (questions 4–5), the centers form a direct equilateral triangle or a single point: Napoleon’s theorem.

**15.** The inward apex is $b + \eu^{+\iu\pi/3}(c - b)$, and the computation of question 11 with $\eu^{+\iu\pi/3} = \frac{1 +
\iu\sqrt3}2$ swaps $\alpha$ and $\beta$: $m_a = \beta b + \alpha
c$, $m_b = \beta c + \alpha a$, $m_c = \beta a + \alpha b$. Using the analogous identities $b + j^2c + ja = j\,Q$ and $c + j^2a + jb
= j^2 Q$ for $Q = a + j^2b + jc$:

$$
m_a + j^2 m_b + j m_c
= \beta(b + j^2 c + ja) + \alpha(c + j^2 a + jb)
= (\beta j + \alpha j^2)\, Q = j(\beta + \alpha j)\,Q = 0 ,
$$

since $\alpha j = -\beta$ (question 14). So the inner centers satisfy the *indirect* equilateral criterion: equilateral with the opposite orientation (or a point).

**16.** By question 5 applied to the triple $(n_a, n_b, n_c)$ (which satisfies the direct criterion), two centers coincide iff all three do; and all three coincide iff *both* criteria hold, i.e. iff additionally $n_a + j^2 n_b + j n_c = 0$. Compute as in question 13, with the identities $b + j^2c + ja = jQ$, $c + j^2a +
jb = j^2Q$:

$$
n_a + j^2 n_b + j n_c
= \alpha\,jQ + \beta\,j^2 Q = j(\alpha + \beta j)\,Q .
$$

Directly: $\beta j = \frac{(3 -
\iu\sqrt3)(-1 + \iu\sqrt3)}{12} = \frac{-3 + 3\iu\sqrt3 +
\iu\sqrt3 + 3}{12} = \frac{\iu\sqrt3}3$, so $\alpha + \beta j =
\frac{3 + \iu\sqrt3 + 2\iu\sqrt3}6 = \frac{1 + \iu\sqrt3}2 \neq
0$. Hence the outer Napoleon triangle degenerates iff $Q = a +
j^2b + jc = 0$, i.e. iff $(a, b, c)$ is an indirect equilateral triangle — in that case the “outward” constructions all point into the triangle’s circumscribed region and share one center.

**17.** With $a = 0$, $b = 1$, $c = \iu$:

$$
n_a = \alpha + \beta\iu = \frac{(3 + \sqrt3)(1 + \iu)}6,
\qquad
n_b = \alpha\iu = \frac{-\sqrt3 + 3\iu}6,
\qquad
n_c = \beta = \frac{3 - \iu\sqrt3}6 .
$$

Squared sides: $n_a - n_b = \frac{(3 + 2\sqrt3) +
\iu\sqrt3}6$ gives $\abs{n_a - n_b}^2 = \frac{(3 + 2\sqrt3)^2 +
3}{36} = \frac{24 + 12\sqrt3}{36} = \frac{2 + \sqrt3}3$; $n_b -
n_c = \frac{(3 + \sqrt3)(-1 + \iu)}6$ gives $\abs{n_b - n_c}^2 =
\frac{2(3 + \sqrt3)^2}{36} = \frac{24 + 12\sqrt3}{36}$; $n_c - n_a
= \frac{-\sqrt3 - \iu(3 + 2\sqrt3)}6$ gives the same value again. All three squared sides equal $\frac{2 + \sqrt3}3$: equilateral, as promised.

**18.** Expand:

$$
(a - b)(c - d) + (a - d)(b - c)
= (ac - ad - bc + bd) + (ab - ac - bd + cd)
= ab - ad - bc + cd ,
$$

and $(a - c)(b - d) = ab - ad - bc + cd$: equal.

**19.** Take moduli in question 18 and apply the triangle inequality ([Proposition 3.2](#prop-b1-complex-rules) (4)):

$$
AC \cdot BD = \abs{(a-b)(c-d) + (a-d)(b-c)}
\leq \abs{a-b}\,\abs{c-d} + \abs{a-d}\,\abs{b-c}
= AB \cdot CD + AD \cdot BC ,
$$

with equality iff the two summands lie on a common ray from $0$ (the equality case of the triangle inequality).

**20.** Half-angle factorization ([Method 3.11](#met-b1-complex-trig) (4)): $\eu^{\iu\theta} - \eu^{\iu\varphi} = 2\iu\,\sin\frac{\theta -
\varphi}2\;\eu^{\iu(\theta + \varphi)/2}$, and taking moduli kills the unimodular factors: $\abs{\eu^{\iu\theta} - \eu^{\iu\varphi}}
= 2\,\abs{\sin\frac{\theta - \varphi}2}$.

**21.** With $p = \eu^{\iu\theta}$, $\theta \in
\intoo{2\pi/3}{4\pi/3}$, question 20 gives

$$
PA = 2\,\abs{\sin\tfrac\theta2},
\quad
PB = 2\,\abs{\sin\bigl(\tfrac\theta2 - \tfrac\pi3\bigr)},
\quad
PC = 2\,\abs{\sin\bigl(\tfrac\theta2 - \tfrac{2\pi}3\bigr)} .
$$

On the arc, $\frac\theta2 \in \intoo{\pi/3}{2\pi/3}$: then $\sin\frac\theta2 > 0$; $\frac\theta2 - \frac\pi3 \in
\intoo0{\pi/3}$, so the second sine is positive; $\frac\theta2 -
\frac{2\pi}3 \in \intoo{-\pi/3}0$, so the third is negative and $PC = 2\sin\bigl(\frac{2\pi}3 - \frac\theta2\bigr)$. Sum to product:

$$
\sin\Bigl(\frac\theta2 - \frac\pi3\Bigr) +
\sin\Bigl(\frac{2\pi}3 - \frac\theta2\Bigr)
= 2\,\sin\frac\pi6\,\cos\Bigl(\frac\theta2 - \frac\pi2\Bigr)
= \sin\frac\theta2 ,
$$

so $PB + PC = 2\sin\frac\theta2 = PA$: van Schooten’s theorem.

**22.** At $p = -1$: $PA = 2$, $PB = PC = 1$ and all sides of the equilateral triangle have length $\sqrt3$, so both sides of Ptolemy read $2\sqrt3$. Algebraically, using $-1 - j^2 = j$ and $1 + j = -j^2$:

$$
(a - b)(p - c) = (1 - j)\,(-1 - j^2) = (1 - j)j = j - j^2
= \iu\sqrt3 ,
$$

$$
(a - c)(b - p) = (1 - j^2)(j + 1) = 1 + j - j^2 - j^3
= j - j^2 = \iu\sqrt3 .
$$

The two terms are equal, so their ratio is $1 \in \R_{>0}$: the equality case of question 19, matching $PA
\cdot BC = PB \cdot AC + PC \cdot AB$ exactly.

**23.** For $(a, b, c) = (1, j, j^2)$: $n_a = \alpha j +
\beta j^2 = -\beta + \beta j^2$ (question 14) $= \beta(j^2 - 1)$; numerically $\beta(j^2 - 1) = \frac{(3 - \iu\sqrt3)}6 \cdot
\bigl(-\frac32 - \iu\frac{\sqrt3}2\bigr) = -1$. Likewise $n_b =
\alpha j^2 + \beta = -j$ and $n_c = \alpha + \beta j = -j^2$. The outer Napoleon triangle of $(1, j, j^2)$ is $(-1, -j, -j^2)$: the original triangle reflected through its centroid $0$ — same size, half-turn rotated. An equilateral triangle is a fixed shape of the Napoleon construction, not a shrinking limit.

**24.** (i) Multiplication by a unimodular number as rotation built the apexes and centers (questions 2–3, 11) and converted circle distances into sines (question 20). (ii) The group structure and the shape invariant justified normalizing the circumcircle to the unit circle and the triangle to $(1, j, j^2)$ in question 21, via the equivariance of question 10. (iii) The half-angle factorization powered both question 20 and the sum-to-product step finishing van Schooten.

**25.** The dictionary converts geometric [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) into polynomial identities in which every hypothesis is an equation: what it buys is mechanization — Napoleon’s theorem reduced to $\alpha j + \beta = 0$, one line of arithmetic in $\Q(\iu\sqrt3)$; what it costs is geometric visibility — the computation certifies but does not show *why* the centers close up into an equilateral triangle. A fair answer is that the identity explains the theorem’s *inevitability* (it holds identically in $a, b, c$, so no cleverness of configuration is involved), while the picture explains its *content*. The same dictionary returns in matrix form when rotations become the orthogonal $2 \times 2$ matrices of [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) and [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid), where “multiplication by $\eu^{\iu\theta}$” is the prototype of a linear isometry.