---
title: "Complex Numbers"
book: "High School Mathematics"
subject: math
language: en
chapter: 28
exercises: 10
source: https://one-course.com/books/math/2/en/chapter/28-complex-numbers
---

# Chapter 28 — Complex Numbers

The [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x^2 = -1$ has no real solution. Enlarging $\R$ with a single new number $\iu$ whose square is $-1$ produces the field $\C$ of [complex numbers](#def-g12-complex-def), in which *every* polynomial [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) has solutions — and whose arithmetic encodes the geometry of the plane: translations, rotations and scalings become additions and multiplications.

## 28.1 Algebraic form

**Definition 28.1 (Complex numbers).**

The set of *complex numbers* is

$$
\C = \{a + \iu b : a, b \in \R\},
$$

where $\iu$ is a symbol subject to the single rule $\iu^2 = -1$; addition and multiplication are performed as for real polynomials in $\iu$, reducing $\iu^2$ to $-1$. The reals $a$ and $b$ are the *real part* $\Rea(z)$ and *imaginary part* $\Ima(z)$ of $z = a + \iu b$; this expression is the *algebraic form* of $z$, and it is unique: $a + \iu b = a' + \iu b'$ if and only if $a = a'$ and $b = b'$.

**Example 28.2.**

$(2 + 3\iu)(1 - \iu) = 2 - 2\iu + 3\iu - 3\iu^2 = 5 + \iu$.

**Definition 28.3 (Conjugate).**

The *conjugate* of $z = a + \iu b$ is $\conj{z} = a - \iu b$.

**Proposition 28.4 (Properties of conjugation).**

For all $z, w \in \C$:

$$
\conj{z + w} = \conj z + \conj w, \quad
\conj{zw} = \conj z\,\conj w, \quad
\conj{\conj z} = z, \quad
z + \conj z = 2\Rea(z), \quad
z - \conj z = 2\iu\,\Ima(z),
$$

and $z\conj z = a^2 + b^2 \geq 0$ for $z = a + \iu b$. Moreover $z \in \R$ if and only if $z = \conj z$.

**Proof.** All are direct computations from the [algebraic forms](#def-g12-complex-def); for instance $z\conj z = (a + \iu b)(a - \iu b) = a^2 - (\iu b)^2 = a^2 + b^2$. ∎

**Method 28.5 (Dividing complex numbers).**

To write a quotient in [algebraic form](#def-g12-complex-def), multiply numerator and denominator by the [conjugate](#def-g12-complex-conjugate) of the denominator:

$$
\frac{1}{2 + \iu} = \frac{2 - \iu}{(2+\iu)(2-\iu)} = \frac{2 - \iu}{5}
= \frac25 - \frac15\iu .
$$

Every nonzero [complex number](#def-g12-complex-def) therefore has an inverse: $\C$ is a *field*.

## 28.2 The complex plane, modulus and argument

**Definition 28.6 (Affix, modulus).**

In the plane with [orthonormal](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system), the point $M(a, b)$ is the *[image](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function)* of $z = a + \iu b$, and $z$ is the *affix* of $M$. The *modulus* of $z$ is the distance from $M$ to the origin:

$$
\abs{z} = \sqrt{a^2 + b^2} = \sqrt{z\conj z}.
$$

![The point M(a,b) of affix z: the modulus z is the length OM, and the conjugate z is the reflection of z in the real axis.](https://one-course.com/images/onecourse/chapters/math-2/g12-complex/fig-0baf1f3783e3.svg)

*The point $M(a,b)$ of [affix](#def-g12-complex-modulus) $z$: the [modulus](#def-g12-complex-modulus) $\abs z$ is the length $OM$, and the [conjugate](#def-g12-complex-conjugate) $\conj z$ is the reflection of $z$ in the real axis.*

**Proposition 28.7 (Properties of the modulus).**

For $z, w \in \C$:

$$
\abs{zw} = \abs z\,\abs w, \qquad
\abs{\tfrac zw} = \tfrac{\abs z}{\abs w}\ (w \neq 0), \qquad
\abs{\conj z} = \abs z, \qquad
\abs{z + w} \leq \abs{z} + \abs{w} \ \text{(triangle inequality)}.
$$

Moreover $\abs{z_B - z_A}$ is the distance between the points of [affixes](#def-g12-complex-modulus) $z_A$ and $z_B$.

**Proof.** $\abs{zw}^2 = zw\,\conj{zw} = z\conj z\, w \conj w = \abs z^2 \abs w^2$ gives the first; the second and third are similar. For the triangle inequality:

$$
\abs{z+w}^2 = (z+w)(\conj z + \conj w)
= \abs z^2 + \abs w^2 + 2\Rea(z \conj w)
\leq \abs z^2 + \abs w^2 + 2\abs{z\conj w}
= \bigl(\abs z + \abs w\bigr)^2,
$$

using $\Rea(u) \leq \abs u$. The distance statement is the Pythagorean formula applied to the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $z_B - z_A$. ∎

**Definition 28.8 (Argument, trigonometric and exponential forms).**

Let $z \neq 0$. An *argument* of $z$, written $\Arg(z)$, is any measure $\theta$ of the angle from the positive real axis to the ray $OM$; it is defined up to $2k\pi$. Writing $r = \abs z$,

$$
z = r(\cos\theta + \iu\sin\theta)
\qquad\text{(trigonometric form)}.
$$

Introducing the notation

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

this becomes the *exponential form* $z = r\,\eu^{\iu\theta}$.

![Exponential form: z is determined by its distance r to the origin and the angle from the positive real axis.](https://one-course.com/images/onecourse/chapters/math-2/g12-complex/fig-b5b1d1679901.svg)

*[Exponential form](#def-g12-complex-argument): $z$ is determined by its distance $r$ to the origin and the angle $\theta$ from the positive real axis.*

**Theorem 28.9.**

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

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

Consequently, for $z = r\eu^{\iu\theta}$ and $z' = r'\eu^{\iu\theta'}$ nonzero:

$$
zz' = rr'\,\eu^{\iu(\theta+\theta')},
\qquad
\frac{z}{z'} = \frac{r}{r'}\,\eu^{\iu(\theta - \theta')} :
$$

*moduli multiply, [arguments](#def-g12-complex-argument) add.*

**Proof.** The first identity *is* the pair of addition formulas ([Proposition 24.3](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#prop-g12-trigo-addition)):

$$
(\cos\theta + \iu\sin\theta)(\cos\theta' + \iu\sin\theta')
= (\cos\theta\cos\theta' - \sin\theta\sin\theta')
+ \iu(\sin\theta\cos\theta' + \cos\theta\sin\theta').
$$

The rest follows by direct computation. ∎

**Corollary 28.10 (De Moivre’s formula).**

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

**Proof.** Induction on $n \geq 0$ using [Theorem 28.9](#thm-g12-complex-expform); extend to $n < 0$ by taking inverses. ∎

**Method 28.11 (Switching between forms).**

From algebraic to exponential: compute $r = \abs z$, then find $\theta$ with $\cos\theta = \frac ar$, $\sin\theta = \frac br$ (locate the correct quadrant before invoking an inverse [cosine](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#def-g12-trigo-cossin)). From exponential to algebraic: expand $r\cos\theta + \iu\, r\sin\theta$. Use the [algebraic form](#def-g12-complex-def) for sums, the [exponential form](#def-g12-complex-argument) for products, powers and quotients.

**Example 28.12.**

$z = 1 + \iu$: $\abs z = \sqrt2$ and $\cos\theta = \sin\theta =
\frac{1}{\sqrt2}$, so $\theta = \frac\pi4$ and $z = \sqrt2\,\eu^{\iu\pi/4}$. Hence $z^{20} = 2^{10}\,\eu^{5\iu\pi} = -1024$.

## 28.3 Second-degree equations and roots of unity

**Theorem 28.13 (Quadratic equations with real coefficients).**

Let $a, b, c \in \R$, $a \neq 0$, and $\Delta = b^2 - 4ac$. The [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $az^2 + bz + c = 0$ has solutions in $\C$:

$$
\begin{aligned}
&\Delta > 0:\ z = \frac{-b \pm \sqrt{\Delta}}{2a} \ (\text{two real}); \qquad
\Delta = 0:\ z = \frac{-b}{2a} \ (\text{double}); \\
&\Delta < 0:\ z = \frac{-b \pm \iu\sqrt{-\Delta}}{2a} \ (\text{two conjugate}).
\end{aligned}
$$

**Proof.** Complete the square: $az^2 + bz + c = a\left(\left(z + \frac{b}{2a}\right)^2 -
\frac{\Delta}{4a^2}\right)$. If $\Delta < 0$, then $\frac{\Delta}{4a^2} = \left(\iu\frac{\sqrt{-\Delta}}{2a}\right)^2$, and the difference of squares factors as usual. ∎

**Definition 28.14 (Roots of unity).**

For $n \geq 1$, the *$n$-th roots of unity* are the solutions of $z^n = 1$.

**Theorem 28.15.**

The [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $z^n = 1$ has exactly $n$ solutions:

$$
\omega_k = \eu^{2\iu k\pi/n}, \qquad k = 0, 1, \dots, n-1 .
$$

Their [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) are the vertices of a regular $n$-gon inscribed in the unit circle, and for $n \geq 2$ their sum is $0$.

**Proof.** Each $\omega_k$ satisfies $\omega_k^n = \eu^{2\iu k\pi} = 1$, and they are pairwise distinct since their [arguments](#def-g12-complex-argument) lie in $\intco{0}{2\pi}$. Conversely, if $z^n = 1$ then $\abs z^n = 1$, so $\abs z = 1$ (positive real) and $z = \eu^{\iu\theta}$ with $n\theta \equiv 0 \pmod{2\pi}$, *i.e.* $\theta = \frac{2k\pi}{n}$; reducing $k$ modulo $n$ lands among the $\omega_k$. The vertices are equally spaced by angle $\frac{2\pi}{n}$: a regular $n$-gon. Finally, with $\omega = \omega_1$, $\omega \neq 1$ and the geometric sum gives

$$
\sum_{k=0}^{n-1}\omega_k = \sum_{k=0}^{n-1}\omega^k
= \frac{\omega^n - 1}{\omega - 1} = 0 . \qedhere
$$

∎

![The fifth roots of unity, = 2 π/5: the vertices of a regular pentagon inscribed in the unit circle ().](https://one-course.com/images/onecourse/chapters/math-2/g12-complex/fig-613555d80b0f.svg)

*The fifth [roots of unity](#def-g12-complex-rootsofunity), $\omega = \eu^{2\iu\pi/5}$: the vertices of a regular pentagon inscribed in the unit circle ([Exercise 28.10](#exo-g12-complex-10)).*

**Example 28.16.**

The cube roots of unity are $1$, $j = \eu^{2\iu\pi/3} =
-\frac12 + \iu\frac{\sqrt3}{2}$ and $\conj j = j^2$, satisfying $1 + j + j^2 = 0$.

## 28.4 Complex numbers and plane geometry

**Proposition 28.17 (Geometric dictionary).**

Let $A, B, C, D$ be points of [affixes](#def-g12-complex-modulus) $a, b, c, d$ ($a \neq b$, $c \neq d$).

1. The translation by the [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) of [affix](#def-g12-complex-modulus) $t$ is $z \mapsto z + t$ .
2. The rotation of center $\omega$ ( [affix](#def-g12-complex-modulus) ) and angle $\theta$ is $z \mapsto \omega + \eu^{\iu\theta}(z - \omega)$ .
3. The scaling (homothety) of center $\omega$ and ratio $k \in \R^*$ is $z \mapsto \omega + k(z - \omega)$ .
4. $\dfrac{d - c}{b - a}$ has [modulus](#def-g12-complex-modulus) $\dfrac{CD}{AB}$ and [argument](#def-g12-complex-argument) the angle between the [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) $\vect{AB}$ and $\vect{CD}$ . In particular, $AB \perp CD$ if and only if $\frac{d-c}{b-a}$ is purely imaginary, and $A, B, C$ are [collinear](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-collinear) ( $C \ne A$ , $C\ne B$ ) if and only if $\frac{c-a}{b-a} \in \R$ .

**Proof.** (1) and (3) restate the coordinate formulas. For (2), the map $w \mapsto \eu^{\iu\theta} w$ multiplies moduli by $1$ and adds $\theta$ to [arguments](#def-g12-complex-argument): it is the rotation of angle $\theta$ about the origin; conjugating by the translation sending $\omega$ to $0$ gives the general center. For (4), moduli and [arguments](#def-g12-complex-argument) of quotients subtract ([Theorem 28.9](#thm-g12-complex-expform)). ∎

## 28.5 Exercises

**Exercise 28.1 ★.**

Write in [algebraic form](#def-g12-complex-def): $(3 - 2\iu)(1 + 4\iu)$, $\dfrac{2 + \iu}{1 - \iu}$, $\iu^{2027}$.

**Solution of Exercise 28.1.**

$(3-2\iu)(1+4\iu) = 3 + 12\iu - 2\iu - 8\iu^2 = 11 + 10\iu$.

$\dfrac{2+\iu}{1-\iu} = \dfrac{(2+\iu)(1+\iu)}{(1-\iu)(1+\iu)}
= \dfrac{1 + 3\iu}{2} = \dfrac12 + \dfrac32\iu$.

$\iu^{2027}$: since $\iu^4 = 1$ and $2027 = 4 \times 506 + 3$, $\iu^{2027} = \iu^3 = -\iu$.

**Exercise 28.2 ★.**

Solve in $\C$: (a) $z^2 - 4z + 13 = 0$; (b) $z^2 + z + 1 = 0$. Check in each case that the two solutions are [conjugates](#def-g12-complex-conjugate) and compute their product.

**Solution of Exercise 28.2.**

*(a)* $\Delta = 16 - 52 = -36$: $z = \dfrac{4 \pm 6\iu}{2} = 2 \pm 3\iu$. [Conjugate](#def-g12-complex-conjugate) pair; product $= 4 + 9 = 13 = \frac ca$. *(b)* $\Delta = 1 - 4 = -3$: $z = \dfrac{-1 \pm \iu\sqrt3}{2}$ (the cube roots of unity $j$ and $\conj j$); product $= \frac{1 + 3}{4} = 1$.

**Exercise 28.3 ★.**

Put in [exponential form](#def-g12-complex-argument): $z_1 = -1 + \iu$, $z_2 = \sqrt3 - \iu$, and compute $\dfrac{z_1}{z_2}$ in exponential then [algebraic form](#def-g12-complex-def). Deduce the exact value of $\cos\dfrac{11\pi}{12}$.

**Solution of Exercise 28.3.**

$\abs{z_1} = \sqrt2$, [argument](#def-g12-complex-argument) $\frac{3\pi}{4}$ (second quadrant): $z_1 = \sqrt2\,\eu^{3\iu\pi/4}$. $\abs{z_2} = 2$, [argument](#def-g12-complex-argument) $-\frac\pi6$: $z_2 = 2\,\eu^{-\iu\pi/6}$. Hence

$$
\frac{z_1}{z_2} = \frac{\sqrt2}{2}\,\eu^{\iu\left(\frac{3\pi}{4} + \frac{\pi}{6}\right)}
= \frac{\sqrt2}{2}\,\eu^{11\iu\pi/12}.
$$

Algebraically,

$$
\frac{z_1}{z_2} = \frac{-1+\iu}{\sqrt3 - \iu}
= \frac{(-1+\iu)(\sqrt3+\iu)}{4}
= \frac{-\sqrt3 - 1 + \iu(\sqrt3 - 1)}{4}.
$$

Identifying [real parts](#def-g12-complex-def): $\frac{\sqrt2}{2}\cos\frac{11\pi}{12} = \frac{-\sqrt3-1}{4}$, so

$$
\cos\frac{11\pi}{12} = -\frac{\sqrt6 + \sqrt2}{4}.
$$

**Exercise 28.4 ★★.**

Describe geometrically the set of points $M$ of [affix](#def-g12-complex-modulus) $z$ such that: (a) $\abs{z - 2} = \abs{z + \iu}$; (b) $\abs{z - 1 - \iu} = 3$; (c) $\dfrac{z - 1}{z + 1}$ is purely imaginary.

**Solution of Exercise 28.4.**

*(a)* $\abs{z - 2} = \abs{z - (-\iu)}$ means $M$ is equidistant from $A(2, 0)$ and $B(0, -1)$: the *perpendicular bisector* of $[AB]$.

*(b)* Distance $3$ from the point $(1,1)$: the *circle* of center $1 + \iu$ and radius $3$.

*(c)* By [Proposition 28.17](#prop-g12-complex-geometry)(4), $\frac{z-1}{z+1}$ purely imaginary means the [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) from $A(1,0)$ to $M$ and from $B(-1,0)$ to $M$ are [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal): $M$ sees the segment $[AB]$ under a right angle. The set is the *circle of diameter $[AB]$* (the unit circle) *minus the points $A$ and $B$* themselves.

**Exercise 28.5 ★★.**

Using De Moivre’s formula and the binomial theorem, express $\cos 3\theta$ as a polynomial in $\cos\theta$, and $\sin 3\theta$ in terms of $\sin\theta$.

**Solution of Exercise 28.5.**

De Moivre: $\cos3\theta + \iu\sin3\theta = (\cos\theta + \iu\sin\theta)^3$. [Expanding](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand) by the binomial theorem with $c = \cos\theta$, $s = \sin\theta$:

$$
(c + \iu s)^3 = c^3 + 3\iu c^2 s - 3 c s^2 - \iu s^3
= (c^3 - 3cs^2) + \iu(3c^2 s - s^3).
$$

Identifying parts and using $s^2 = 1 - c^2$, $c^2 = 1 - s^2$:

$$
\cos3\theta = 4\cos^3\theta - 3\cos\theta,
\qquad
\sin3\theta = 3\sin\theta - 4\sin^3\theta .
$$

**Exercise 28.6 ★★.**

*(Linearization.)* Using $\cos\theta = \frac{\eu^{\iu\theta} + \eu^{-\iu\theta}}{2}$, linearize $\cos^3\theta$ (write it as a combination of $\cos k\theta$), and deduce $\displaystyle\int_0^{\pi/2}\cos^3\theta\,\dd\theta$.

**Solution of Exercise 28.6.**

$$
\cos^3\theta = \left(\frac{\eu^{\iu\theta} + \eu^{-\iu\theta}}{2}\right)^{\!3}
= \frac{\eu^{3\iu\theta} + 3\eu^{\iu\theta} + 3\eu^{-\iu\theta} + \eu^{-3\iu\theta}}{8}
= \frac{\cos3\theta + 3\cos\theta}{4}.
$$

Hence

$$
\int_0^{\pi/2}\cos^3\theta\,\dd\theta
= \frac14\left[\frac{\sin3\theta}{3} + 3\sin\theta\right]_0^{\pi/2}
= \frac14\left(-\frac13 + 3\right) = \frac{2}{3}.
$$

(Consistent with $W_3 = \frac23$ in [Exercise 25.9](https://one-course.com/books/math/2/en/chapter/25-integration#exo-g12-integ-9), since $\int_0^{\pi/2}\cos^3 = \int_0^{\pi/2}\sin^3$ by the symmetry $\theta \mapsto \frac\pi2 - \theta$.)

**Exercise 28.7 ★★.**

Let $A$ and $B$ have [affixes](#def-g12-complex-modulus) $a = 1 + \iu$ and $b = 3 + 2\iu$. Determine the two points $C$ making the triangle $ABC$ equilateral, as the [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) of $B$ under the rotations of center $A$ and angles $\pm\frac{\pi}{3}$.

**Solution of Exercise 28.7.**

$c = a + \eu^{\pm\iu\pi/3}(b - a)$ with $b - a = 2 + \iu$ and $\eu^{\pm\iu\pi/3} = \frac12 \pm \iu\frac{\sqrt3}{2}$:

$$
\eu^{\iu\pi/3}(2+\iu) = \left(\tfrac12 + \iu\tfrac{\sqrt3}{2}\right)(2+\iu)
= \left(1 - \tfrac{\sqrt3}{2}\right) + \iu\left(\tfrac12 + \sqrt3\right),
$$

so $c_1 = \left(2 - \frac{\sqrt3}{2}\right) +
\iu\left(\frac32 + \sqrt3\right)$, and similarly with the angle $-\frac\pi3$: $c_2 = \left(2 + \frac{\sqrt3}{2}\right) + \iu\left(\frac32 - \sqrt3\right)$.

**Exercise 28.8 ★★★.**

Solve $z^4 = -4$ in $\C$ and plot the solutions. Factor $z^4 + 4$ into two quadratic polynomials with real coefficients.

**Solution of Exercise 28.8.**

$-4 = 4\,\eu^{\iu\pi}$, so the solutions of $z^4 = -4$ are

$$
z_k = \sqrt2\,\eu^{\iu\left(\frac\pi4 + \frac{k\pi}{2}\right)},
\quad k = 0, 1, 2, 3,
$$

*i.e.* $1 + \iu$, $-1 + \iu$, $-1 - \iu$, $1 - \iu$: the vertices of a square of circumradius $\sqrt 2$. Grouping [conjugate](#def-g12-complex-conjugate) [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant),

$$
z^4 + 4 = \bigl(z - (1{+}\iu)\bigr)\bigl(z - (1{-}\iu)\bigr)
\bigl(z + (1{-}\iu)\bigr)\bigl(z + (1{+}\iu)\bigr)
= (z^2 - 2z + 2)(z^2 + 2z + 2).
$$

**Exercise 28.9 ★★★.**

For $\theta \in \intoo{0}{2\pi}$ and $n \in \N$, compute

$$
S = 1 + \cos\theta + \cos 2\theta + \dots + \cos n\theta
$$

by summing the geometric progression $\sum_{k=0}^n \eu^{\iu k\theta}$ and taking [real parts](#def-g12-complex-def). (Answer: $S = \dfrac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}
\cos\frac{n\theta}{2}$.)

**Solution of Exercise 28.9.**

With $q = \eu^{\iu\theta} \neq 1$:

$$
\sum_{k=0}^{n} \eu^{\iu k\theta} = \frac{\eu^{\iu(n+1)\theta} - 1}{\eu^{\iu\theta} - 1}
= \frac{\eu^{\iu(n+1)\theta/2}}{\eu^{\iu\theta/2}}
\cdot\frac{\eu^{\iu(n+1)\theta/2} - \eu^{-\iu(n+1)\theta/2}}{\eu^{\iu\theta/2} - \eu^{-\iu\theta/2}}
= \eu^{\iu n\theta/2}\,
\frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}},
$$

using $\eu^{\iu\alpha} - \eu^{-\iu\alpha} = 2\iu\sin\alpha$ (the *half-angle [factoring](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand)*). Taking [real parts](#def-g12-complex-def):

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

**Exercise 28.10 ★★★.**

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

1. Justify that $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ .
2. Let $u = \omega + \omega^4$ and $v = \omega^2 + \omega^3$ . Show that $u + v = -1$ and $uv = -1$ , and deduce that $u = \frac{-1+\sqrt5}{2}$ .
3. Conclude that $\cos\dfrac{2\pi}{5} = \dfrac{\sqrt5 - 1}{4}$ .

**Solution of Exercise 28.10.**

*1.* $\omega$ is a $5$-th [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) of unity different from $1$, so the sum of all five [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) vanishes ([Theorem 28.15](#thm-g12-complex-rootsofunity)), which reads $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$.

*2.* $u + v = \omega + \omega^2 + \omega^3 + \omega^4 = -1$ by point 1. For the product, using $\omega^5 = 1$:

$$
uv = (\omega + \omega^4)(\omega^2 + \omega^3)
= \omega^3 + \omega^4 + \omega^6 + \omega^7
= \omega^3 + \omega^4 + \omega + \omega^2 = -1 .
$$

So $u, v$ are the [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) of $X^2 + X - 1 = 0$: $\left\{\frac{-1+\sqrt5}{2}, \frac{-1-\sqrt5}{2}\right\}$. Now $u = \omega + \conj\omega = 2\cos\frac{2\pi}{5} > 0$ (since $\frac{2\pi}{5} < \frac{\pi}{2}$), so $u = \frac{-1 + \sqrt5}{2}$.

*3.* $\cos\frac{2\pi}{5} = \frac u2 = \frac{\sqrt5 - 1}{4}$.

## 28.6 Problem: The pentagon in the mirror of unity

**Problem 28.1.**

Weekend problem — the fifth roots of unity compute $\cos 72^\circ$ exactly, the golden ratio signs the result, and multiplication turns out to be rotation

Euclid could construct the regular pentagon but the reason *why* it yields to ruler and compass — while the humble $20^\circ$ angle does not — stayed hidden for two thousand years. It hides in this chapter: the five fifth [roots of unity](#def-g12-complex-rootsofunity) ([Theorem 28.15](#thm-g12-complex-rootsofunity)) sum to zero, and that one line of algebra squeezes $\cos 72^\circ$ out of a quadratic [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) — square roots only, hence constructible, and signed, of course, by the golden ratio. Around this jewel: fluency, de Moivre’s shortcut to trigonometry, and multiplication unmasked as geometry.

**Part I — Fluency.**

1. Compute $(2 + \iu)(3 - \iu)$ , $\dfrac{1 + \iu}{1 - \iu}$ ( [Method 28.5](#met-g12-complex-division) ), and $\abs{3 + 4\iu}$ .
2. Put in [exponential form](#def-g12-complex-argument) : $1 + \iu$ ; $-2$ ; $1 - \iu\sqrt3$ .
3. Compute $(1 + \iu)^8$ using the [exponential form](#def-g12-complex-argument) .
4. Solve $z^2 = \iu$ , then $z^2 - 2z + 5 = 0$ ( [Theorem 28.13](#thm-g12-complex-quadratic) ).
5. Geometric dictionary ( [Proposition 28.17](#prop-g12-complex-geometry) ): describe the map $z \mapsto \iu z$ , and the set $\abs{z - (1 + \iu)} = 2$ .

**Part II — Five [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant), one pentagon.** Let $\omega = \eu^{2\iu\pi/5}$ and consider $1, \omega, \omega^2, \omega^3, \omega^4$.

6. Check these are exactly the solutions of $z^5 = 1$ , and describe the figure they draw in the plane.
7. Prove that their sum is zero. (Geometric progression — or multiply the sum by $1 - \omega$ .)
8. Take [real parts](#def-g12-complex-def) and deduce $$1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0 .$$
9. Set $c = \cos\frac{2\pi}{5}$ and use the double-angle formula for $\cos\frac{4\pi}{5}$ to convert question 8 into a quadratic [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) for $c$; solve it and conclude $$\cos 72^\circ = \frac{\sqrt5 - 1}{4} .$$
10. Verify numerically, then let the golden ratio sign its work: show $c = \frac{1}{2\varphi}$ where $\varphi = \frac{1 + \sqrt5}{2}$ ( [Problem 2.1](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#pb-g10-algebra-1) ) — and state the famous geometric echo: in a regular pentagon, the diagonal is $\varphi$ times the side.

**Part III — De Moivre’s shortcuts.**

11. The construction verdict: $\cos 72^\circ$ needs only square roots (question 9), so the pentagon is ruler-and-compass constructible; $\cos 20^\circ$ satisfies an unbreakable cubic ( [Problem 24.1](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#pb-g12-trigo-1) ), so the $60^\circ$ angle cannot be trisected. State the emerging criterion (square roots buildable, cube roots not), whose full theory — Gauss, age 19, and the 17-gon — lives in the university volumes.
12. Solve $z^4 = -4$ by [exponential form](#def-g12-complex-argument) and recover [Exercise 28.8](#exo-g12-complex-8) ’s factorization $z^4 + 4 = (z^2 - 2z + 2)(z^2 + 2z + 2)$ by pairing [conjugate](#def-g12-complex-conjugate) [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) .
13. Expand $(\cos\theta + \iu\sin\theta)^3$ with the binomial theorem and de Moivre, and re-derive $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ in three lines — compare with the addition-formula route of [Problem 24.1](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#pb-g12-trigo-1) .
14. Cross-check with [Exercise 28.9](#exo-g12-complex-9) : for $\theta = \frac{2\pi}{5}$ and $n = 4$ , what does the sum $1 + \cos\theta + \dots + \cos 4\theta$ equal, and why is that consistent with question 8?
15. Compute the three cube roots of $8\iu$ in [algebraic form](#def-g12-complex-def) .

**Part IV — Multiplication is geometry.**

16. Describe completely the map $z \mapsto (1 + \iu)z$ : by what angle does it rotate, by what factor does it scale, and what does it do to the unit square?
17. Prove $\abs{z_1 z_2} = \abs{z_1}\,\abs{z_2}$ with [conjugates](#def-g12-complex-conjugate) ( [Proposition 28.7](#prop-g12-complex-modulus) ), then compute $(1+\iu)^n$ for $n = 1, 2, 3, 4$ and describe the spiral the powers trace.
18. The number $\mathrm{j} = \eu^{2\iu\pi/3}$ : show $1 + \mathrm{j} + \mathrm{j}^2 = 0$ , and verify on the triangle $\left(1, \mathrm{j}, \mathrm{j}^2\right)$ the classical criterion: $a + \mathrm{j} b + \mathrm{j}^2 c = 0$ holds for (one orientation of) an equilateral triangle $abc$ .
19. The fundamental theorem of algebra (admitted: d’Alembert–Gauss): every polynomial factors completely over $\C$ . Deduce the real-world corollary — every real polynomial factors into real linear and quadratic pieces — and point to question 12 as an instance.
20. Finale — $\C$ ’s portrait, one sentence each: numbers become points; multiplication becomes rotation-and-scaling; the [roots of unity](#def-g12-complex-rootsofunity) become regular polygons; de Moivre turns powers into trigonometric identities; and algebraic closure guarantees that no [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) ever needs a bigger world. Coda: which two celebrated numbers met in question 10?

**Solution of Problem 28.1.**

**1.** $(2 + \iu)(3 - \iu) = 6 - 2\iu + 3\iu + 1 =
7 + \iu$. $\frac{1 + \iu}{1 - \iu} =
\frac{(1 + \iu)^2}{2} = \frac{2\iu}{2} = \iu$. $\abs{3 + 4\iu} = 5$.

**2.** $1 + \iu = \sqrt2\,\eu^{\iu\pi/4}$; $-2 = 2\eu^{\iu\pi}$; $1 - \iu\sqrt3 = 2\eu^{-\iu\pi/3}$.

**3.** $(1 + \iu)^8 = \left(\sqrt2\right)^8
\eu^{8\iu\pi/4} = 16\,\eu^{2\iu\pi} = 16$.

**4.** $z^2 = \eu^{\iu\pi/2}$: $z = \pm\eu^{\iu\pi/4} = \pm\frac{\sqrt2}{2}(1 + \iu)$. And $\Delta = 4 - 20 = -16$: $z = 1 \pm 2\iu$.

**5.** $z \mapsto \iu z$: rotation by a quarter turn about the origin. $\abs{z - (1 + \iu)} = 2$: the circle of center $1 + \iu$ and radius $2$.

**6.** $\left(\omega^k\right)^5 = \eu^{2\iu\pi k} = 1$, and the five points $\eu^{2\iu k\pi/5}$ are distinct: all five solutions of $z^5 = 1$. They sit on the unit circle at equal angles of $72^\circ$: a regular pentagon with one vertex at $1$.

**7.** $S = 1 + \omega + \dots + \omega^4$ satisfies $(1 - \omega)S = 1 - \omega^5 = 0$, and $\omega \neq 1$: $S = 0$. (The pentagon’s balance point is its center.)

**8.** $\omega^4 = \overline\omega$ and $\omega^3 = \overline{\omega^2}$: pairing [conjugates](#def-g12-complex-conjugate), the [real part](#def-g12-complex-def) of the sum reads $1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0$.

**9.** $\cos\frac{4\pi}{5} = 2c^2 - 1$, so $1 + 2c + 4c^2 - 2 = 0$, i.e. $4c^2 + 2c - 1 = 0$: $c = \frac{-1 + \sqrt5}{4}$ (the positive [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant), since $72^\circ$ is acute): $\cos 72^\circ = \frac{\sqrt5 - 1}{4}$.

**10.** Numerically $c \approx 0.30902$: matches the calculator. And $\frac{1}{2\varphi} = \frac{1}{1 + \sqrt5} =
\frac{\sqrt5 - 1}{4}$ (rationalize): $c = \frac{1}{2\varphi}$. The pentagon is golden through and through: its diagonal cuts its side in the ratio $\varphi$ — the pentagram’s endless self-similarity.

**11.** Square roots are compass-constructible (the half-circle machine of the Middle School volume’s geometric [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean)), so a number built from rationals by square roots alone — like $\cos 72^\circ$ — is constructible; cube roots are not, and $\cos 20^\circ$ needs one: the pentagon yields, trisection resists. Gauss’s criterion settles every regular polygon — his 17-gon at nineteen made him choose mathematics.

**12.** $-4 = 4\eu^{\iu\pi}$: the fourth [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) have [modulus](#def-g12-complex-modulus) $\sqrt2$ and [arguments](#def-g12-complex-argument) $\frac{\pi}{4} + k\frac{\pi}{2}$: the four numbers $\pm 1 \pm \iu$. Pairing [conjugates](#def-g12-complex-conjugate): $(z - 1 - \iu)(z - 1 + \iu) = z^2 - 2z + 2$ and $(z + 1 - \iu)(z + 1 + \iu) = z^2 + 2z + 2$: the factorization of [Exercise 28.8](#exo-g12-complex-8).

**13.** $(\cos\theta + \iu\sin\theta)^3 = \cos^3\theta +
3\iu\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta -
\iu\sin^3\theta$; by de Moivre this equals $\cos 3\theta + \iu \sin 3\theta$. [Real parts](#def-g12-complex-def): $\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta)
= 4\cos^3\theta - 3\cos\theta$ — and the [imaginary parts](#def-g12-complex-def) hand over $\sin 3\theta$ for free.

**14.** It is the [real part](#def-g12-complex-def) of $1 + \omega + \dots + \omega^4 = 0$: the sum equals $0$ — exactly question 8, seen through [Exercise 28.9](#exo-g12-complex-9)’s kernel.

**15.** $8\iu = 8\eu^{\iu\pi/2}$: [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) $2\eu^{\iu\pi/6} = \sqrt3 + \iu$; $2\eu^{5\iu\pi/6} = -\sqrt3 + \iu$; $2\eu^{3\iu\pi/2} = -2\iu$.

**16.** $1 + \iu = \sqrt2\,\eu^{\iu\pi/4}$: the map rotates the plane by $45^\circ$ and scales it by $\sqrt2$. The unit square becomes a tilted square of side $\sqrt2$ — doubled in area, spun an eighth of a turn.

**17.** $\abs{z_1 z_2}^2 = z_1 z_2 \overline{z_1 z_2} =
z_1\overline{z_1}\, z_2\overline{z_2} =
\abs{z_1}^2\abs{z_2}^2$: take square roots. Powers of $1 + \iu$: $1 + \iu$, $2\iu$, $-2 + 2\iu$, $-4$: each a $45^\circ$ turn further and $\sqrt2$ times longer — an outward spiral through moduli $\sqrt2, 2, 2\sqrt2, 4, \dots$

**18.** $\mathrm{j}$ is a third [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) of unity distinct from $1$, so (question 7’s [argument](#def-g12-complex-argument) with $n = 3$) $1 + \mathrm{j} + \mathrm{j}^2 = 0$. Criterion on the standard triangle: $a + \mathrm{j}b + \mathrm{j}^2 c = 1 +
\mathrm{j}\cdot\mathrm{j} + \mathrm{j}^2 \cdot \mathrm{j}^2 =
1 + \mathrm{j}^2 + \mathrm{j}^4 = 1 + \mathrm{j}^2 +
\mathrm{j} = 0$: satisfied, as the equilateral triangle $\left(1, \mathrm{j}, \mathrm{j}^2\right)$ demands.

**19.** Over $\C$, a real polynomial splits into linear factors; its non-real [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) come in [conjugate](#def-g12-complex-conjugate) pairs ([conjugate](#def-g12-complex-conjugate) the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation)), and each pair multiplies into a real quadratic $z^2 - 2\operatorname{Re}(z_0)\,z + \abs{z_0}^2$: hence the real factorization into degree-$1$ and degree-$2$ pieces — question 12 performed it on $z^4 + 4$.

**20.** Points; rotation-and-scaling; regular polygons; trigonometric identities by expansion; and a closed world where every [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of degree $n$ has its $n$ [roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant). Coda: in question 10, $\pi$’s pentagon shook hands with the golden ratio — Euclid’s two most famous guests, introduced at last by a sum of five arrows equal to zero.