CommeUnJeu · L1 MPSI

Complex numbers

⌚ ~123 min ▢ 15 blocks ✓ 67 exercises ➣ Prerequisites : Complex Numbers: Algebraic Approach, Complex Numbers: Geometrical Approach

The equation $x^2 = -1$ has no real solution: in $\mathbb{R}$, every square is non-negative. The story of complex numbers is the story of insisting --- against this obstacle --- that such an equation should have a solution, and following the consequences. We admit the existence of a set $\mathbb{C}$ containing $\mathbb{R}$, equipped with an element $i$ such that $i^2 = -1$, in which every element writes uniquely as $a + ib$ with $a, b \in \mathbb{R}$, and on which addition and multiplication extend those of $\mathbb{R}$. (The construction of $\mathbb{C}$ from $\mathbb{R}^2$ is hors programme; we use this set as a given.)
The answer is $\mathbb{C}$, the field of complex numbers. Each $z \in \mathbb{C}$ writes uniquely $z = a + ib$ with $a, b \in \mathbb{R}$, so $\mathbb{C}$ is, as a set, a copy of the plane $\mathbb{R}^2$. But $\mathbb{C}$ is much more: it is a field, the algebraic structure of $\mathbb{R}$ enriched with a multiplication that has a geometric meaning. Multiplying by $e^{i\theta}$ is rotating by an angle $\theta$ around the origin; multiplying by a positive real is dilating. Together, the two operations encode the direct similitudes of the plane.
The plan has four movements. We start by setting up the algebraic machinery: the algebraic form $a+ib$, the conjugate $\conjugate{z}$, the modulus $|z|$, and the triangle inequality. Next, we introduce the trigonometric form $z = \rho\, e^{i\theta}$, Moivre's and Euler's formulas, and the complex exponential. Then we solve equations: square roots, second-degree equations in $\mathbb{C}$, and the $n$-th roots of unity. Finally, we cash in the geometry: affixes, alignments, orthogonality, and the direct similitudes $z \mapsto az + b$.
A constant subtext: algebra and geometry are two sides of the same coin. Every algebraic identity in $\mathbb{C}$ is a geometric statement about the plane; every geometric move is an algebraic operation on a complex number. Once that bridge is in place, problems from one side become tools for the other.

I The field $\mathbb{C}$ and its first invariants

We start by laying down the four objects on which everything else rests: the algebraic form $a+ib$, the conjugate, the modulus, and the triangle inequality. Each is defined with a one-line algebraic formula and a one-line geometric reading. Spend the time to internalize both sides --- it is the entire payoff of the chapter.

I.1 Algebraic form$\virgule$ real and imaginary parts

We adjoin to $\mathbb{R}$ a new symbol $i$ satisfying $i^2 = -1$, and consider the set of all $a + ib$ with $a, b \in \mathbb{R}$. Addition and multiplication are extended from $\mathbb{R}$ in the obvious way --- distributivity, associativity, commutativity --- with the single new rule $i \cdot i = -1$. The resulting set is denoted $\mathbb{C}$, and the writing $a+ib$ turns out to be unique: each complex number corresponds to exactly one pair $(a, b)$.

Definition — The set $\mathbb{C}$ and the algebraic form

The set of complex numbers is $$ \mathbb{C} = \{ a + ib \ \mid \ a, b \in \mathbb{R} \} $$ where $i$ is a symbol satisfying $\textcolor{colordef}{i^2 = -1}$.
Addition and multiplication are defined, for $a, b, c, d \in \mathbb{R}$, by $$ (a+ib) + (c+id) = (a+c) + i(b+d) \qquad (a+ib)(c+id) = (ac - bd) + i(ad + bc). $$
For $z = a + ib$, the writing $a+ib$ is the algebraic form (or Cartesian form) of $z$.
$a$ is the real part of $z$, denoted $\textcolor{colordef}{\operatorname{Re}(z)}$;
$b$ is the imaginary part of $z$, denoted $\textcolor{colordef}{\operatorname{Im}(z)}$.

Both $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ are real numbers.

Example

For $z = 3 + 2i$ and $w = 1 - i$: $$ z + w = 4 + i \qquad z\cdot w = (3 + 2i)(1 - i) = 3 - 3i + 2i - 2i^2 = 5 - i. $$ The real and imaginary parts read off the algebraic form: $\operatorname{Re}(z) = 3$, $\operatorname{Im}(z) = 2$.

Example

A pure imaginary number is a $z$ with $\operatorname{Re}(z) = 0$, that is, of the form $z = ib$. Examples: $i$, $-3i$, $i\pi$. Conversely, a real number $a$ is the special case $b = 0$, so $\mathbb{R} \subset \mathbb{C}$.

Proposition — Equality and uniqueness of algebraic form

For $z = a+ib$ and $w = c+id$ in $\mathbb{C}$ with $a, b, c, d \in \mathbb{R}$: $$ z = w \ \iff \ a = c \ \text{and} \ b = d. $$ In particular, the writing $a+ib$ of $z \in \mathbb{C}$ is unique: $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ are well-defined.

Proof

$z = w$ rewrites as $(a-c) + i(b-d) = 0$. We must show $a = c$ and $b = d$. Suppose $b \ne d$: then $i = (c-a)/(b-d)$ would be a real number. But $i^2 = -1 < 0$, while every real square is non-negative --- contradiction. So $b = d$, and then $a - c = 0$, i.e. $a = c$. The converse is immediate.

Example

The complex number $z = a + ib$ is identified with the point of the plane of coordinates $(a, b)$. The horizontal axis carries the real numbers $\mathbb{R}$; the vertical axis carries the pure imaginary numbers $i\mathbb{R}$. This identification is called the Argand plane.

Method — Compute the algebraic form of a quotient

To put $z = (a+ib)/(c+id)$ with $c+id \ne 0$ in algebraic form, multiply numerator and denominator by $c - id$: $$ \frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{(c+id)(c-id)} = \frac{(ac+bd) + i(bc - ad)}{c^2 + d^2}. $$ The denominator becomes the real number $c^2 + d^2$, and the numerator expands by distributivity. This trick will reappear constantly.

Example

Compute $\dfrac{1+i}{1-i}$ in algebraic form.

Answer

$$ \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{2i}{2} = i. $$ So the quotient is the pure imaginary $i$.

Skills to practice

Computing in algebraic form

I.2 Conjugate

The conjugate is the simplest of all complex transformations: change the sign of the imaginary part. Algebraically it is a one-line definition; geometrically it is the reflection across the real axis. Its key feature is that it is compatible with addition and multiplication: $\overline{z+w} = \bar z + \bar w$ and $\overline{zw} = \bar z \cdot \bar w$. From this single fact almost all algebraic manipulations of complex expressions follow. (We will see later, in the chapter on algebraic structures, that this makes conjugation a field automorphism.)

Definition — Conjugate

For $z = a + ib \in \mathbb{C}$ with $a, b \in \mathbb{R}$, the conjugate of $z$ is $$ \textcolor{colordef}{\conjugate{z}} = a - ib. $$ In other words, conjugation flips the sign of the imaginary part.

Example

$\conjugate{3 + 2i} = 3 - 2i$.
$\conjugate{-1 + i\sqrt{3}} = -1 - i\sqrt{3}$.
$\conjugate{5} = 5$ (a real number is its own conjugate).
$\conjugate{i} = -i$ (a pure imaginary flips sign).

Example

The point $\conjugate{z}$ is the reflection of $z$ across the real axis.

Proposition — Algebraic properties of the conjugate

For all $z, w \in \mathbb{C}$:

$\textcolor{colorprop}{\conjugate{\conjugate{z}} = z}$ (involution).
$\textcolor{colorprop}{\conjugate{z + w} = \conjugate{z} + \conjugate{w}}$.
$\textcolor{colorprop}{\conjugate{z \cdot w} = \conjugate{z} \cdot \conjugate{w}}$.
If $w \ne 0$, $\textcolor{colorprop}{\conjugate{1/w} = 1/\conjugate{w}}$ and $\textcolor{colorprop}{\conjugate{z/w} = \conjugate{z}/\conjugate{w}}$.
For all $n \in \mathbb{Z}$ (with $z \ne 0$ if $n < 0$), $\textcolor{colorprop}{\conjugate{z^n} = \conjugate{z}^n}$.

Proof

Write $z = a + ib$ and $w = c + id$ with $a, b, c, d \in \mathbb{R}$.

Involution. $\conjugate{\conjugate{z}} = \conjugate{a - ib} = a + ib = z$.
Addition. $\conjugate{z+w} = \conjugate{(a+c) + i(b+d)} = (a+c) - i(b+d) = (a-ib) + (c-id) = \conjugate{z} + \conjugate{w}$.
Multiplication. $z\cdot w = (ac - bd) + i(ad+bc)$, so $\conjugate{zw} = (ac-bd) - i(ad+bc)$. On the other hand $\conjugate{z}\conjugate{w} = (a-ib)(c-id) = (ac - bd) - i(ad+bc)$. The two coincide.
Inverse. For $w \ne 0$, $w \cdot (1/w) = 1$, so $\conjugate{w}\cdot \conjugate{1/w} = \conjugate{1} = 1$, giving $\conjugate{1/w} = 1/\conjugate{w}$. The quotient case follows by combining the multiplication and inverse rules.
Power. For $n \ge 0$ by induction; for $n < 0$ apply the inverse rule to $z^{-n}$.

Proposition — Real part$\virgule$ imaginary part$\virgule$ real and pure imaginary characterization

For all $z \in \mathbb{C}$: $$ \textcolor{colorprop}{\operatorname{Re}(z) = \frac{z + \conjugate{z}}{2}} \qquad \textcolor{colorprop}{\operatorname{Im}(z) = \frac{z - \conjugate{z}}{2i}}. $$ Consequently: $$ \textcolor{colorprop}{z \in \mathbb{R} \ \iff \ \conjugate{z} = z} \qquad \textcolor{colorprop}{z \in i\mathbb{R} \ \iff \ \conjugate{z} = -z}. $$

Proof

With $z = a + ib$, $z + \conjugate{z} = 2a$ and $z - \conjugate{z} = 2ib$, hence the two formulas. The two equivalences then follow from $\operatorname{Im}(z) = 0 \iff z \in \mathbb{R}$ and $\operatorname{Re}(z) = 0 \iff z \in i\mathbb{R}$.

Example

Show that for all $z \in \mathbb{C}$, $z\conjugate{z}$ is a real non-negative number.

Answer

With $z = a + ib$ $$ z\conjugate{z} = (a+ib)(a-ib) = a^2 - (ib)^2 = a^2 + b^2 \ge 0. $$ This identity will be the cornerstone of the next subsection.

Skills to practice

Manipulating conjugates

I.3 Modulus

The modulus of a complex number is its distance from the origin in the Argand plane. Algebraically it is $\sqrt{a^2 + b^2}$, the Pythagorean length of $(a, b)$. Together with the conjugate, it satisfies the master identity $z\conjugate{z} = |z|^2$ --- the engine that turns most calculations on $|\cdot|$ into algebra.

Definition — Modulus

For $z = a + ib \in \mathbb{C}$ with $a, b \in \mathbb{R}$, the modulus of $z$ is the non-negative real number $$ \textcolor{colordef}{|z|} = \sqrt{a^2 + b^2}. $$ For $z \in \mathbb{R}$, $|z|$ coincides with the absolute value, justifying the same notation.

Example

$|3 + 4i| = \sqrt{9 + 16} = 5$.
$|-1 + i\sqrt{3}| = \sqrt{1 + 3} = 2$.
$|i| = 1$.
$|0| = 0$.

Example

The modulus is the length of the vector from the origin to the point $z$ --- the Pythagorean diagonal of the rectangle with sides $|\operatorname{Re}(z)|$ and $|\operatorname{Im}(z)|$.

Proposition — Master identity and basic properties

For all $z, w \in \mathbb{C}$:

$\textcolor{colorprop}{|z|^2 = z \conjugate{z}}$ (master identity).
$\textcolor{colorprop}{|z| \ge 0}$ and $\textcolor{colorprop}{|z| = 0 \iff z = 0}$.
$\textcolor{colorprop}{|\conjugate{z}| = |z|}$, $\textcolor{colorprop}{|-z| = |z|}$.
$\textcolor{colorprop}{|z\,w| = |z|\cdot|w|}$ (multiplicativity).
If $w \ne 0$, $\textcolor{colorprop}{|z/w| = |z|/|w|}$.
For all $n \in \mathbb{Z}$ (with $z \ne 0$ if $n < 0$), $\textcolor{colorprop}{|z^n| = |z|^n}$.
$\textcolor{colorprop}{|\operatorname{Re}(z)| \le |z|}$ and $\textcolor{colorprop}{|\operatorname{Im}(z)| \le |z|}$.

Proof

Master identity. For $z = a+ib$, $z\conjugate{z} = a^2 + b^2 = |z|^2$.
Positivity and separation. $|z| = \sqrt{a^2+b^2} \ge 0$, with equality iff $a = b = 0$ iff $z = 0$.
Conjugate and opposite. $|\conjugate{z}|^2 = a^2 + b^2 = |z|^2$; same for $-z$.
Multiplicativity. $|zw|^2 = (zw)\conjugate{zw} = z\conjugate{z}\cdot w\conjugate{w} = |z|^2|w|^2$. Taking square roots gives $|zw| = |z||w|$.
Quotient. Apply multiplicativity to $z = (z/w)\cdot w$: $|z| = |z/w|\cdot |w|$, hence $|z/w| = |z|/|w|$.
Power. Induction on $n \ge 0$ using multiplicativity; extend to $n < 0$ via the quotient rule.
Bounds. $a^2 \le a^2 + b^2$, so $|a| \le |z|$. Same for $|b|$.

Method — Compute a modulus efficiently

To compute $|z|$ for a complex expression $z$:

if $z$ is given in algebraic form $a + ib$, use $|z| = \sqrt{a^2 + b^2}$;
if $z$ is given as a product or quotient, exploit multiplicativity: $|zw| = |z||w|$, $|z/w| = |z|/|w|$;
if $z$ is a power $z = w^n$, use $|z| = |w|^n$;
to avoid square roots until the end, work with $|z|^2 = z\conjugate{z}$.

The third strategy turns $|(1+i)^{20}|$ into $|1+i|^{20} = (\sqrt 2)^{20} = 2^{10} = 1024$, in one line.

Example

Compute $\left| \dfrac{(1+i)^5}{2 - i} \right|$.

Answer

By multiplicativity and the power rule: $$ \left| \frac{(1+i)^5}{2-i} \right| = \frac{|1+i|^5}{|2-i|} = \frac{(\sqrt 2)^5}{\sqrt 5} = \frac{4\sqrt 2}{\sqrt 5} = \frac{4\sqrt{10}}{5}. $$

Example

Determine the set of $z \in \mathbb{C}$ with $|z - 1| = |z + i|$.

Answer

Squaring (both sides are non-negative) and using $|w|^2 = w\conjugate{w}$: $$ \begin{aligned} |z-1|^2 = |z+i|^2 \ &\iff \ (z-1)(\conjugate{z}-1) = (z+i)(\conjugate{z}-i) \\ &\iff \ z\conjugate{z} - z - \conjugate{z} + 1 = z\conjugate{z} - iz + i\conjugate{z} + 1 \\ &\iff \ -(z + \conjugate{z}) = i(\conjugate{z} - z) \\ &\iff \ -2\operatorname{Re}(z) = 2\operatorname{Im}(z) \\ &\iff \ \operatorname{Re}(z) = -\operatorname{Im}(z). \end{aligned} $$ The set is the line of equation $y = -x$ in the Argand plane --- which is, geometrically, the set of points equidistant from $1$ and $-i$ (the perpendicular bisector of the segment $[1, -i]$).

Proposition — Circles and disks via the modulus

For $a \in \mathbb{C}$ and $r \ge 0$, the modulus $|z - a|$ measures the distance from the point of affix $z$ to the point of affix $a$. Hence:

$\textcolor{colorprop}{\{z \in \mathbb{C} : |z - a| = r\}}$ is the circle of center $a$ and radius $r$.
$\textcolor{colorprop}{\{z \in \mathbb{C} : |z - a| \le r\}}$ is the closed disk of center $a$ and radius $r$.
$\textcolor{colorprop}{\{z \in \mathbb{C} : |z - a| < r\}}$ is the open disk of center $a$ and radius $r$.

Example

Describe the set $\{z \in \mathbb{C} : |z - (1 + i)| = 2\}$.

Answer

Circle of center $1 + i$ and radius $2$.

Skills to practice

Computing moduli and characterizing loci

I.4 Triangle inequality

The triangle inequality is the modular analogue of the geometric statement « in a triangle, each side is at most the sum of the other two ». It is the most-used inequality of the chapter --- and one of the rare inequalities in $\mathbb{C}$, since $\mathbb{C}$ has no order. Its equality case is also worth remembering: it characterizes when two complex numbers are positively colinear.

Theorem — Triangle inequality

For all $z, w \in \mathbb{C}$: $$ \textcolor{colorprop}{\big| |z| - |w| \big| \le |z + w| \le |z| + |w|}. $$ The right-hand inequality is an equality if and only if $z$ and $w$ are positively colinear: there exists $\lambda \ge 0$ such that $w = \lambda z$, or $z = 0$ (which makes the equality automatic).

Proof

Right-hand inequality. Both sides are non-negative; we square. By the master identity: $$ |z + w|^2 = (z+w)(\conjugate{z}+\conjugate{w}) = |z|^2 + z\conjugate{w} + \conjugate{z}w + |w|^2 = |z|^2 + 2\operatorname{Re}(z\conjugate{w}) + |w|^2. $$ Now $\operatorname{Re}(z\conjugate{w}) \le |z\conjugate{w}| = |z||w|$, so $$ |z+w|^2 \le |z|^2 + 2|z||w| + |w|^2 = (|z|+|w|)^2. $$ Taking square roots: $|z+w| \le |z|+|w|$.
Equality case. Equality holds iff $\operatorname{Re}(z\conjugate{w}) = |z\conjugate{w}|$, which means $z\conjugate{w}$ is a non-negative real number. If $z = 0$, the conclusion is automatic. Otherwise, multiplying $z\conjugate{w} \ge 0$ by $1/(z\conjugate{z}) = 1/|z|^2 > 0$ preserves the sign: $\conjugate{w}/\conjugate{z} \ge 0$, hence $w/z \ge 0$, i.e. $w = \lambda z$ with $\lambda \ge 0$ --- the form stated in the theorem.
Left-hand inequality. Apply the right-hand inequality to $z = (z+w) + (-w)$: $|z| \le |z+w| + |-w| = |z+w| + |w|$, so $|z| - |w| \le |z+w|$. By symmetry, $|w| - |z| \le |z+w|$, hence $\big||z|-|w|\big| \le |z+w|$.

Example

The vector $z+w$ is the third side of the triangle with vertices $0$, $z$, $z+w$; the other two sides have lengths $|z|$ and $|w|$. The triangle inequality is the geometric statement « one side $\le$ sum of the other two ».

Example

For $|z| \le 2$, bound $|z^2 + 1|$.
Triangle inequality: $|z^2 + 1| \le |z^2| + 1 = |z|^2 + 1 \le 5$. The bound is reached at $z = 2$ where $|z^2 + 1| = 5$.

Example

For $z \in \mathbb{C}$ with $|z| = 1$, show $|1 + z| \le 2$ with equality iff $z = 1$.
By the triangle inequality, $|1 + z| \le |1| + |z| = 1 + 1 = 2$. Equality requires positive colinearity of $1$ and $z$, i.e. $z = \lambda \cdot 1$ with $\lambda \ge 0$; combined with $|z| = 1$, $\lambda = 1$, so $z = 1$.

Method — Prove a modular inequality

A typical inequality problem in $\mathbb{C}$ has the form « show $|f(z)| \le g(|z|) $ ». The standard playbook:

replace $|f(z)|^2$ by $f(z)\conjugate{f(z)}$ when convenient;
apply the triangle inequality $|a + b| \le |a| + |b|$ or its lower version $|a + b| \ge ||a| - |b||$;
apply multiplicativity $|ab| = |a||b|$ to break down products;
use the bounds $|\operatorname{Re}(w)| \le |w|$, $|\operatorname{Im}(w)| \le |w|$;
finally, simplify using the hypothesis on $|z|$.

For sharp inequalities, identify the equality case (positive colinearity) and check whether it is attained under the hypotheses.

Skills to practice

Bounding moduli with the triangle inequality

II Trigonometric form and complex exponential

We now switch viewpoint. Instead of writing $z$ as a sum of real and imaginary parts, we read it as a polar coordinate: a modulus $\rho$ and an angle $\theta$. The notation $e^{i\theta}$ for $\cos\theta + i\sin\theta$ then unifies trigonometry and exponentiation: products become sums of angles, powers become multiples of angles, the formulas $\cos(n\theta)$ and $\sin(n\theta)$ unfold from the binomial. This second viewpoint is what makes complex numbers a tool of geometry and analysis.

II.1 The unit circle $\mathbb{U}$ and the imaginary exponential

The complex numbers of modulus $1$ form a circle: the unit circle of the Argand plane. Each such number is determined by its angle from the positive real axis. We introduce the notation $e^{i\theta}$ for $\cos\theta + i\sin\theta$ --- a notation that proves itself by a single algebraic identity, $e^{i\theta}\,e^{i\varphi} = e^{i(\theta + \varphi)}$, equivalent to the addition formulas of cosine and sine.

Definition — The unit circle $\mathbb{U}$

The set of complex numbers of modulus $1$ is denoted $$ \textcolor{colordef}{\mathbb{U}} = \{ z \in \mathbb{C} \ \mid \ |z| = 1 \}. $$ Geometrically, $\mathbb{U}$ is the unit circle of the Argand plane.

Definition — Imaginary exponential

For $\theta \in \mathbb{R}$, we set $$ \textcolor{colordef}{e^{i\theta}} = \cos\theta + i\sin\theta. $$ The number $e^{i\theta}$ belongs to $\mathbb{U}$ since $|e^{i\theta}|^2 = \cos^2\theta + \sin^2\theta = 1$.

Example

$e^{i\cdot 0} = 1$.
$e^{i\pi/2} = i$.
$e^{i\pi} = -1$ (Euler's identity, often written $e^{i\pi} + 1 = 0$).
$e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3) = \dfrac{1}{2} + i\dfrac{\sqrt 3}{2}$.
$e^{i\pi/4} = \dfrac{\sqrt 2}{2}(1 + i)$.

Example

The point $e^{i\theta}$ sits on the unit circle at angle $\theta$ (measured counter-clockwise from the positive real axis). When $\theta$ runs over $[0, 2\pi)$, $e^{i\theta}$ describes the entire unit circle exactly once.

Proposition — Functional equation of the imaginary exponential

For all $\theta, \varphi \in \mathbb{R}$: $$ \textcolor{colorprop}{e^{i\theta}\cdot e^{i\varphi} = e^{i(\theta + \varphi)}} \qquad \textcolor{colorprop}{\frac{1}{e^{i\theta}} = \conjugate{e^{i\theta}} = e^{-i\theta}}. $$ In particular, $e^{i\theta} = e^{i\varphi} \iff \theta \equiv \varphi \,[2\pi]$.

Proof

Functional equation. Compute the product directly: $$ \begin{aligned} e^{i\theta}\cdot e^{i\varphi} &= (\cos\theta + i\sin\theta)(\cos\varphi + i\sin\varphi) \\ &= (\cos\theta\cos\varphi - \sin\theta\sin\varphi) + i(\cos\theta\sin\varphi + \sin\theta\cos\varphi) \\ &= \cos(\theta + \varphi) + i\sin(\theta + \varphi) \\ &= e^{i(\theta + \varphi)}. \end{aligned} $$ The third equality uses the addition formulas for $\cos$ and $\sin$.
Inverse and conjugate. $\conjugate{e^{i\theta}} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta) = e^{-i\theta}$. And $e^{i\theta}\cdot e^{-i\theta} = e^{i\cdot 0} = 1$, so $1/e^{i\theta} = e^{-i\theta}$.
Periodicity. $e^{i\theta} = e^{i\varphi}$ rewrites $e^{i(\theta - \varphi)} = 1$, i.e. $\cos(\theta - \varphi) = 1$ and $\sin(\theta - \varphi) = 0$, which means $\theta - \varphi \in 2\pi\mathbb{Z}$.

Example

Compute $e^{i\pi/3}\cdot e^{i\pi/6}$.

Answer

By the functional equation, $e^{i\pi/3}\cdot e^{i\pi/6} = e^{i(\pi/3 + \pi/6)} = e^{i\pi/2} = i$.

Skills to practice

Computing with $e^{i\theta}$

II.2 Moivre and Euler formulas

Moivre's formula --- $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ --- is the prize: it gives, in one stroke, $\cos n\theta$ and $\sin n\theta$ as polynomials in $\cos\theta, \sin\theta$ (just expand with the binomial theorem). Its dual, Euler's formulas, give cosine and sine themselves as expressions in $e^{i\theta}$. Together they form a back-and-forth dictionary that converts trigonometric identities into algebraic ones and vice versa.

Theorem — Moivre's formula

For all $\theta \in \mathbb{R}$ and all $n \in \mathbb{Z}$: $$ \textcolor{colorprop}{(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)} $$ or equivalently $\textcolor{colorprop}{(e^{i\theta})^n = e^{in\theta}}$.

Proof

Case $n \ge 0$, induction on $n$. For $n = 0$: both sides equal $1$. Assume $(e^{i\theta})^n = e^{in\theta}$. Then $(e^{i\theta})^{n+1} = (e^{i\theta})^n\cdot e^{i\theta} = e^{in\theta}\cdot e^{i\theta} = e^{i(n+1)\theta}$ by the functional equation. Heredity holds.
Case $n < 0$. Let $n = -m$ with $m > 0$. Then $(e^{i\theta})^n = 1/(e^{i\theta})^m = 1/e^{im\theta} = e^{-im\theta} = e^{in\theta}$.

Proposition — Euler's formulas

For all $\theta \in \mathbb{R}$: $$ \textcolor{colorprop}{\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}} \qquad \textcolor{colorprop}{\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}}. $$

Proof

$e^{i\theta} = \cos\theta + i\sin\theta$ and $e^{-i\theta} = \cos\theta - i\sin\theta$. Adding gives $2\cos\theta$; subtracting gives $2i\sin\theta$.

Example

Compute $(1 + i)^{20}$ via Moivre.

Answer

Write $1 + i = \sqrt 2 \cdot e^{i\pi/4}$. By multiplicativity of moduli and Moivre $$ (1+i)^{20} = (\sqrt 2)^{20} \cdot e^{i\cdot 20\pi/4} = 2^{10} \cdot e^{i \cdot 5\pi} = 1024 \cdot e^{i\pi} = -1024. $$

Example

Linearize $\cos^3\theta$ via Euler.

Answer

$$ \begin{aligned} \cos^3\theta &= \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^3 \\ &= \frac{1}{8}\left( e^{3i\theta} + 3e^{i\theta} + 3e^{-i\theta} + e^{-3i\theta} \right) \\ &= \frac{1}{8}\left( 2\cos 3\theta + 6\cos\theta \right) \\ &= \frac{1}{4}\cos 3\theta + \frac{3}{4}\cos\theta. \end{aligned} $$

Example

Express $\sin(3\theta)$ as a polynomial in $\sin\theta$.

Answer

Moivre with $n = 3$: $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$. Expanding the cube and identifying imaginary parts: $$ (\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta $$ so $\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta = 3(1-\sin^2\theta)\sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta$.

Method — Linearize $\cos^p\theta\sin^q\theta$ via Euler

To express a product $\cos^p\theta\sin^q\theta$ as a linear combination of $\cos(k\theta)$ and $\sin(k\theta)$:

replace $\cos\theta$ by $(e^{i\theta}+e^{-i\theta})/2$ and $\sin\theta$ by $(e^{i\theta}-e^{-i\theta})/(2i)$;
expand both powers using the binomial theorem;
collect terms in $e^{ik\theta}$ and pair $e^{ik\theta}$ with $e^{-ik\theta}$ to recover $2\cos(k\theta)$ or $2i\sin(k\theta)$;
simplify the prefactor.

The procedure always terminates and gives the exact answer; it is the cleanest tool for such problems.

Method — Factor $a\cos t + b\sin t$ as $A\cos(t - \varphi)$

For $(a, b) \ne (0, 0)$, set $A = \sqrt{a^2 + b^2}$. Then $(a/A, b/A)$ is on the unit circle, so there exists a unique $\varphi \in \,]-\pi, \pi]$ such that $\cos\varphi = a/A$ and $\sin\varphi = b/A$. Then $$ a\cos t + b\sin t = A(\cos\varphi\cos t + \sin\varphi\sin t) = A\cos(t - \varphi). $$ Hint for the derivation: $a\cos t + b\sin t = \operatorname{Re}((a - ib)e^{it})$, and $a - ib = A\, e^{-i\varphi}$ by definition of $\varphi$.

Example

Factor $\cos t + \sqrt 3\sin t$.

Answer

$A = \sqrt{1 + 3} = 2$, $\cos\varphi = 1/2$, $\sin\varphi = \sqrt 3/2$, so $\varphi = \pi/3$. Hence $$ \cos t + \sqrt 3\sin t = 2\cos\!\left(t - \frac{\pi}{3}\right). $$

Proposition — Half-angle technique

For $t \in \mathbb{R}$: $$ \textcolor{colorprop}{1 + e^{it} = 2\cos\!\left(\tfrac{t}{2}\right) e^{it/2}}, \qquad \textcolor{colorprop}{1 - e^{it} = -2i\sin\!\left(\tfrac{t}{2}\right) e^{it/2}}. $$ More generally, for $p, q \in \mathbb{R}$: $$ \textcolor{colorprop}{e^{ip} + e^{iq} = 2\cos\!\left(\tfrac{p-q}{2}\right) e^{i(p+q)/2}}, \qquad \textcolor{colorprop}{e^{ip} - e^{iq} = 2i\sin\!\left(\tfrac{p-q}{2}\right) e^{i(p+q)/2}}. $$

Proof

Factor $e^{it/2}$ from $1 + e^{it} = e^{it/2}(e^{-it/2} + e^{it/2}) = e^{it/2} \cdot 2\cos(t/2)$ by Euler. Same for $1 - e^{it} = e^{it/2}(e^{-it/2} - e^{it/2}) = e^{it/2} \cdot (-2i\sin(t/2))$. The general formulas follow by factoring $e^{i(p+q)/2}$ from $e^{ip} \pm e^{iq}$ and applying the special case.

Proposition — Sum-to-product formulas

For $p, q \in \mathbb{R}$, taking real and imaginary parts of the half-angle identities: $$ \textcolor{colorprop}{\cos p + \cos q = 2\cos\!\left(\tfrac{p+q}{2}\right)\cos\!\left(\tfrac{p-q}{2}\right)}, $$ $$ \textcolor{colorprop}{\cos p - \cos q = -2\sin\!\left(\tfrac{p+q}{2}\right)\sin\!\left(\tfrac{p-q}{2}\right)}, $$ $$ \textcolor{colorprop}{\sin p + \sin q = 2\sin\!\left(\tfrac{p+q}{2}\right)\cos\!\left(\tfrac{p-q}{2}\right)}, $$ $$ \textcolor{colorprop}{\sin p - \sin q = 2\cos\!\left(\tfrac{p+q}{2}\right)\sin\!\left(\tfrac{p-q}{2}\right)}. $$

Proposition — Trigonometric sums

For $t \in \mathbb{R}$ with $t \not\equiv 0 \,[2\pi]$ and $n \in \mathbb{N}$: $$ \sum_{k=0}^{n} e^{ikt} = \frac{1 - e^{i(n+1)t}}{1 - e^{it}}. $$ Taking real and imaginary parts and applying the half-angle technique to the quotient: $$ \textcolor{colorprop}{\sum_{k=0}^{n} \cos(kt) = \frac{\sin\!\left(\tfrac{(n+1)t}{2}\right)}{\sin\!\left(\tfrac{t}{2}\right)} \cos\!\left(\tfrac{nt}{2}\right)}, \qquad \textcolor{colorprop}{\sum_{k=0}^{n} \sin(kt) = \frac{\sin\!\left(\tfrac{(n+1)t}{2}\right)}{\sin\!\left(\tfrac{t}{2}\right)} \sin\!\left(\tfrac{nt}{2}\right)}. $$ For $t \equiv 0 \,[2\pi]$: $\sum \cos(kt) = n+1$ and $\sum \sin(kt) = 0$.

Proof

The first identity is the geometric sum (chapter `CalculAlgebrique`) applied to $z = e^{it}$, valid since $z \ne 1$ when $t \not\equiv 0 \,[2\pi]$. For the closed form, factor by half-angle: $$ \frac{1 - e^{i(n+1)t}}{1 - e^{it}} = \frac{-2i\sin\!\left(\tfrac{(n+1)t}{2}\right) e^{i(n+1)t/2}}{-2i\sin\!\left(\tfrac{t}{2}\right) e^{it/2}} = \frac{\sin\!\left(\tfrac{(n+1)t}{2}\right)}{\sin\!\left(\tfrac{t}{2}\right)} e^{int/2}. $$ Take real and imaginary parts of $e^{int/2} = \cos(nt/2) + i\sin(nt/2)$.

Skills to practice

Applying Moivre and linearizing with Euler

II.3 Argument and trigonometric form

Every nonzero complex number $z$ writes uniquely as $z = \rho\, e^{i\theta}$ with $\rho = |z| > 0$ and $\theta$ a real number defined modulo $2\pi$. The pair $(\rho, \theta)$ is the polar coordinate of $z$. The angle $\theta$ is called the argument; making it well-defined requires choosing a representative, and we will adopt the convention $\theta \in \,]-\pi, \pi]$ (the principal argument).

Proposition — Existence and uniqueness of trigonometric form

For all $z \in \mathbb{C}^* = \mathbb{C}\setminus\{0\}$, there exist $\rho > 0$ and $\theta \in \mathbb{R}$ such that $$ \textcolor{colorprop}{z = \rho\, e^{i\theta}}. $$ The number $\rho$ is uniquely determined and equals $|z|$. The number $\theta$ is determined modulo $2\pi$: if $z = \rho_1 e^{i\theta_1} = \rho_2 e^{i\theta_2}$ then $\rho_1 = \rho_2$ and $\theta_1 \equiv \theta_2 \,[2\pi]$.

Proof

Existence. Set $\rho = |z| > 0$. Then $u = z/\rho$ has modulus $1$, so $u = a + ib$ with $a^2 + b^2 = 1$. The pair $(a, b)$ lies on the unit circle of $\mathbb{R}^2$, hence (admitted, from the parametrization of the circle) there exists $\theta \in \mathbb{R}$ such that $a = \cos\theta$ and $b = \sin\theta$. Then $u = \cos\theta + i\sin\theta = e^{i\theta}$, so $z = \rho\, e^{i\theta}$.
Uniqueness of $\rho$. Taking moduli: $\rho_1 = |\rho_1 e^{i\theta_1}| = |z| = \rho_2$.
Uniqueness of $\theta$ modulo $2\pi$. From $\rho_1 = \rho_2 = \rho$, we get $\rho e^{i\theta_1} = \rho e^{i\theta_2}$, i.e. $e^{i\theta_1} = e^{i\theta_2}$, i.e. $e^{i(\theta_1 - \theta_2)} = 1$. By the periodicity of the imaginary exponential established above, $\theta_1 - \theta_2 \in 2\pi\mathbb{Z}$, i.e. $\theta_1 \equiv \theta_2 \,[2\pi]$.

Definition — Argument and trigonometric form

By the existence theorem of the trigonometric form just established, every nonzero $z \in \mathbb{C}$ writes $z = \rho\, e^{i\theta}$ with $\rho > 0$ uniquely determined and $\theta$ determined modulo $2\pi$. We call:

$\rho = |z|$ the modulus of $z$ (already named in the Modulus subsection above);
$\theta$ an argument of $z$, denoted $\arg(z)$ (defined modulo $2\pi$);
the unique representative $\theta \in \,]-\pi, \pi]$ the principal argument;
the writing $z = \rho\, e^{i\theta}$ the trigonometric form (or polar form) of $z$.

The number $z = 0$ has no argument.

Example

For $z = 1 + i\sqrt 3$: $\rho = |z| = \sqrt{1+3} = 2$, and the angle $\theta$ satisfies $\cos\theta = 1/2$, $\sin\theta = \sqrt 3/2$, hence $\theta = \pi/3$. Trigonometric form: $z = 2 e^{i\pi/3}$.

Method — Convert from algebraic to trigonometric form

Starting from $z = a + ib$ with $z \ne 0$:

compute $\rho = |z| = \sqrt{a^2 + b^2}$;
factor: $z = \rho\!\left( \dfrac{a}{\rho} + i\dfrac{b}{\rho} \right)$;
identify $\theta$ from the system $\cos\theta = a/\rho$ and $\sin\theta = b/\rho$ --- both equations are needed (the sign of $\sin\theta$ disambiguates the half-plane).

Avoid $\theta = \arctan(b/a)$ as a starting point: arctan only handles the right half-plane $a > 0$, and even then misses the cases $a = 0$.

Example

Find the trigonometric form of $z = -\sqrt 3 + i$.

Answer

$\rho = \sqrt{3 + 1} = 2$. Then $\cos\theta = -\sqrt 3/2$ and $\sin\theta = 1/2$, so $\theta = 5\pi/6$ (second quadrant). Hence $z = 2\, e^{i\,5\pi/6}$.

Proposition — Argument: algebraic properties

For all $z, w \in \mathbb{C}^*$ and all $n \in \mathbb{Z}$:

$\textcolor{colorprop}{\arg(zw) \equiv \arg(z) + \arg(w) \,[2\pi]}$;
$\textcolor{colorprop}{\arg(z/w) \equiv \arg(z) - \arg(w) \,[2\pi]}$;
$\textcolor{colorprop}{\arg(z^n) \equiv n\arg(z) \,[2\pi]}$;
$\textcolor{colorprop}{\arg(\conjugate{z}) \equiv -\arg(z) \,[2\pi]}$.

Proof

Write $z = \rho_z e^{i\theta_z}$, $w = \rho_w e^{i\theta_w}$.

Product. $zw = \rho_z\rho_w e^{i(\theta_z+\theta_w)}$, so $\arg(zw) \equiv \theta_z + \theta_w \,[2\pi]$.
Quotient. Apply the product rule to $z = (z/w) \cdot w$: $\arg(z) \equiv \arg(z/w) + \arg(w)$, hence $\arg(z/w) \equiv \arg(z) - \arg(w)$.
Power. Induction on $n \ge 0$: for $n = 0$, $\arg(1) \equiv 0$ ; the inductive step uses the product rule. For $n < 0$, write $z^n = 1/z^{-n}$ and apply the quotient rule.
Conjugate. $\conjugate{z} = \rho_z e^{-i\theta_z}$ (since $\conjugate{e^{i\theta}} = e^{-i\theta}$), so $\arg(\conjugate{z}) \equiv -\theta_z \equiv -\arg(z)$.

Example

Multiplying by $z_1 = i = e^{i\pi/2}$ rotates by a quarter-turn: $z_2$ at argument $\pi/4$ is sent to $z_1 z_2$ at argument $3\pi/4$. More generally, multiplying by $\rho\, e^{i\theta}$ is « rotate by $\theta$, then dilate by $\rho$ ». This is the geometric meaning of complex multiplication --- and the foundation of the geometric interpretation in the next section.

Skills to practice

Putting in trigonometric form

II.4 Complex exponential

We extend the exponential to all of $\mathbb{C}$ by combining the two exponentials we know: the real exponential of $\operatorname{Re}(z)$, and the imaginary exponential of $\operatorname{Im}(z)$. The two pieces multiply, and the resulting function $z \mapsto e^z$ inherits the cardinal property of the exponential: $e^{z+w} = e^z e^w$. Solving $e^z = a$ for given $a \in \mathbb{C}^*$ becomes a routine matter of polar coordinates.

Definition — Complex exponential

For $z = x + iy \in \mathbb{C}$ with $x, y \in \mathbb{R}$, we set $$ \textcolor{colordef}{e^z} = e^x \cdot e^{iy} = e^x(\cos y + i\sin y). $$ The function $\exp: \mathbb{C} \to \mathbb{C}$, $z \mapsto e^z$, is called the complex exponential.

Proposition — Properties of the complex exponential

For all $z, w \in \mathbb{C}$:

$\textcolor{colorprop}{e^{z + w} = e^z\cdot e^w}$ (cardinal property).
$\textcolor{colorprop}{e^z \ne 0}$ for all $z \in \mathbb{C}$, and $\textcolor{colorprop}{1/e^z = e^{-z}}$.
$\textcolor{colorprop}{|e^z| = e^{\operatorname{Re}(z)}}$ and $\textcolor{colorprop}{\arg(e^z) \equiv \operatorname{Im}(z) \,[2\pi]}$.
$\textcolor{colorprop}{\conjugate{e^z} = e^{\conjugate{z}}}$.
$\textcolor{colorprop}{e^z = 1 \iff z \in 2i\pi\mathbb{Z}}$. More generally, $\textcolor{colorprop}{e^z = e^w \iff z - w \in 2i\pi\mathbb{Z}}$.

Proof

Write $z = x+iy$, $w = u+iv$.

Cardinal property. $e^z e^w = e^x e^{iy} e^u e^{iv} = e^{x+u} e^{i(y+v)} = e^{(x+u)+i(y+v)} = e^{z+w}$ by the real exponential's morphism property and the imaginary exponential's functional equation.
Non-vanishing. $e^z = 0$ would force $e^x = 0$ (impossible since $|e^{iy}| = 1$). And $e^z e^{-z} = e^0 = 1$ gives $1/e^z = e^{-z}$.
Modulus and argument. $|e^z| = |e^x|\cdot|e^{iy}| = e^x = e^{\operatorname{Re}(z)}$; $\arg(e^z) \equiv y \equiv \operatorname{Im}(z) \,[2\pi]$.
Conjugate. $\conjugate{e^z} = \conjugate{e^x e^{iy}} = e^x e^{-iy} = e^{x - iy} = e^{\conjugate{z}}$.
Periodicity. $e^z = 1$ rewrites $e^x = 1$ and $e^{iy} = 1$, i.e. $x = 0$ and $y \in 2\pi\mathbb{Z}$, i.e. $z \in 2i\pi\mathbb{Z}$. The general case $e^z = e^w$ follows from $e^{z-w} = 1$.

Example

$e^{i\pi} = -1$.
$e^{1 + i\pi/2} = e\cdot e^{i\pi/2} = ei$.
$e^{\ln 2 + i\pi} = 2\cdot (-1) = -2$.
$e^{2 - i\pi/3} = e^2(\cos(-\pi/3) + i\sin(-\pi/3)) = e^2\!\left(\frac{1}{2} - i\frac{\sqrt 3}{2}\right)$.

Example

Solve $e^z = -2$.

Answer

Write the right-hand side in polar form: $-2 = 2\, e^{i\pi}$. We seek $z = x + iy$ with $e^x = 2$ and $e^{iy} = e^{i\pi}$. The first gives $x = \ln 2$; the second gives $y \equiv \pi \,[2\pi]$. Hence $z = \ln 2 + i\pi + 2ik\pi$ for $k \in \mathbb{Z}$.

Skills to practice

Solving exponential equations

III Algebraic equations in $\mathbb{C}$

The original motivation for $\mathbb{C}$ was to make $x^2 = -1$ solvable. We now reap the harvest. In $\mathbb{C}$, every nonzero complex has exactly two opposed square roots, every quadratic equation has exactly two solutions (counted with multiplicity), and more generally every nonzero complex has exactly $n$ distinct $n$-th roots. The proofs are short --- they are read off the trigonometric form --- and each one comes with a vivid geometric picture: the $n$-th roots of unity sit on the unit circle as the vertices of a regular $n$-gon.

III.1 Square roots of a complex number

A square root of $a \in \mathbb{C}$ is a complex $\delta$ such that $\delta^2 = a$. For $a = 0$, the only root is $\delta = 0$. For $a \ne 0$, we will see there are exactly two roots, and they are opposites of each other. Two computational paths exist: trigonometric (read off $\rho$ and $\theta$) and algebraic (solve a small system in real and imaginary parts). Both have their place.

Theorem — Square roots of a complex

Let $a \in \mathbb{C}^*$. The equation $\delta^2 = a$ has exactly two solutions in $\mathbb{C}$, opposite to each other. Concretely, if $a = \rho\, e^{i\theta}$ in trigonometric form, the two square roots are $$ \textcolor{colorprop}{\{\delta : \delta^2 = a\} = \{\sqrt\rho\, e^{i\theta/2},\ -\sqrt\rho\, e^{i\theta/2}\}}. $$ The choice of $\theta$ representative determines the labeling, but the unordered pair is intrinsic. We will write $\delta_1, \delta_2$ for the two roots when a particular labeling is fixed.

Proof

Verification. $\delta_1^2 = \rho\, e^{i\theta} = a$. And $\delta_2^2 = \delta_1^2 = a$. So both are roots.
Exhaustiveness. If $\delta^2 = a$, write $\delta = r\, e^{i\varphi}$ in trigonometric form ($\delta \ne 0$ since $a \ne 0$). Then $r^2 e^{2i\varphi} = \rho\, e^{i\theta}$, so $r^2 = \rho$ (positive real, hence $r = \sqrt\rho$) and $2\varphi \equiv \theta \,[2\pi]$, i.e. $\varphi \equiv \theta/2 \,[\pi]$. So $\varphi$ is either $\theta/2$ or $\theta/2 + \pi$ modulo $2\pi$, giving $\delta = \delta_1$ or $\delta = \delta_2$.

Method — Compute square roots --- trigonometric path

For $a \in \mathbb{C}^*$:

put $a$ in trigonometric form $\rho\, e^{i\theta}$;
the two square roots are $\pm\sqrt\rho\, e^{i\theta/2}$.

This is the cleanest path when $a$ is in $\mathbb{U}$ or has a simple polar form.

Method — Compute square roots --- algebraic path

For $a = \alpha + i\beta$ with $\alpha, \beta \in \mathbb{R}$, seek $\delta = x + iy$ with $\delta^2 = a$. Squaring and identifying real and imaginary parts: $$ x^2 - y^2 = \alpha \qquad 2xy = \beta. $$ Add $|\delta|^2 = |a|$, i.e. $x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}$. The first and the last give $x^2$ and $y^2$ as positive expressions; signs are tied by $2xy = \beta$ (same sign as $\beta$ if $\beta \ne 0$).

Example

Compute the square roots of $a = -2 + 2i\sqrt 3$ (trigonometric path).

Answer

$|a| = \sqrt{4 + 12} = 4$, so $a = 4(\cos\theta + i\sin\theta)$ with $\cos\theta = -1/2$, $\sin\theta = \sqrt 3/2$, hence $\theta = 2\pi/3$. Thus $a = 4\, e^{2i\pi/3}$. The roots are $\pm 2\, e^{i\pi/3} = \pm(1 + i\sqrt 3)$.

Example

Compute the square roots of $a = 3 + 4i$ (algebraic path).

Answer

$|a| = 5$. Seek $\delta = x + iy$ with $x^2 + y^2 = 5$, $x^2 - y^2 = 3$, $2xy = 4$. Adding the first two: $x^2 = 4$; subtracting: $y^2 = 1$. Since $2xy = 4 > 0$, $x$ and $y$ have the same sign. Hence $\delta = \pm(2 + i)$.

Example

The two square roots of $a$ sit on the circle of radius $\sqrt{|a|}$, opposite each other (separated by angle $\pi$). They are read off as the half-angle of $a$.

Skills to practice

Computing square roots

III.2 Quadratic equations in $\mathbb{C}$

The quadratic equation $az^2 + bz + c = 0$ with complex coefficients has the same canonical form as in the real case, but the discriminant $\Delta = b^2 - 4ac$ is now complex. The novelty is that $\Delta$ always has square roots in $\mathbb{C}$ (we just proved it), so the formula $z = (-b \pm \delta)/(2a)$ always works. There is no « no real solution » case; in $\mathbb{C}$ every quadratic has exactly two solutions (counted with multiplicity).

Theorem — Quadratic equation in $\mathbb{C}$

Let $a, b, c \in \mathbb{C}$ with $a \ne 0$. The equation $az^2 + bz + c = 0$ has solutions $$ \textcolor{colorprop}{z = \frac{-b + \delta}{2a} \quad\text{or}\quad z = \frac{-b - \delta}{2a}} $$ where $\delta$ is any square root of $\Delta = b^2 - 4ac$. The two solutions $z_1, z_2$ satisfy the Viète relations: $$ \textcolor{colorprop}{z_1 + z_2 = -\frac{b}{a} \qquad z_1 z_2 = \frac{c}{a}}. $$ The solutions coincide ($z_1 = z_2$) if and only if $\Delta = 0$.

Proof

Multiply by $4a \ne 0$: $$ 4a(az^2 + bz + c) = (2az + b)^2 - (b^2 - 4ac) = (2az + b)^2 - \Delta. $$ Let $\delta$ be a square root of $\Delta$. The equation becomes $(2az + b)^2 = \delta^2$, i.e. $2az + b = \pm \delta$, hence $z = (-b \pm \delta)/(2a)$. Viète: $z_1 + z_2 = -2b/(2a) = -b/a$; $z_1 z_2 = (b^2 - \delta^2)/(4a^2) = (b^2 - \Delta)/(4a^2) = 4ac/(4a^2) = c/a$. Finally $z_1 - z_2 = \delta/a$, and since $a \ne 0$, the two solutions coincide iff $\delta = 0$, i.e. iff $\Delta = \delta^2 = 0$.

Method — Solve a quadratic in $\mathbb{C}$

For $az^2 + bz + c = 0$ with $a \ne 0$:

compute $\Delta = b^2 - 4ac$;
find a square root $\delta$ of $\Delta$ (algebraic or trigonometric path);
write $z = (-b \pm \delta)/(2a)$.

If the coefficients are real and $\Delta$ is a real number, the usual real treatment applies and $\Delta < 0$ gives $\delta = i\sqrt{-\Delta}$, leading to two conjugate solutions.

Example

Solve $z^2 - (3+i)z + (2+i) = 0$.

Answer

$\Delta = (3+i)^2 - 4(2+i) = 9 + 6i - 1 - 8 - 4i = 2i = 2\, e^{i\pi/2}$. A square root is $\delta = \sqrt 2\, e^{i\pi/4} = 1 + i$. Solutions: $z = ((3+i) \pm (1+i))/2$, i.e. $z_1 = 2 + i$ and $z_2 = 1$. Check: $z_1 z_2 = (2+i)(1) = 2 + i = c/a$ $\checkmark$; $z_1 + z_2 = 3 + i = -b/a$ $\checkmark$.

Proposition — Factorization by a root

Let $P$ be a complex polynomial function and $a \in \mathbb{C}$. If $P(a) = 0$, then there exists a complex polynomial function $Q$ such that $$ \textcolor{colorprop}{P(z) = (z - a)\, Q(z) \quad \text{for all } z \in \mathbb{C}}. $$

Proof

Write $P(z) = \sum_{k=0}^{n} c_k z^k$ and use the elementary identity (chapter `CalculAlgebrique`) $$ z^k - a^k = (z - a) \sum_{j=0}^{k-1} z^{k-1-j} a^{j}. $$ Then $P(z) - P(a) = \sum_{k=0}^{n} c_k (z^k - a^k) = (z - a) \sum_{k=1}^{n} c_k \sum_{j=0}^{k-1} z^{k-1-j} a^j$. Since $P(a) = 0$, the left-hand side is $P(z)$ itself. Setting $Q(z)$ to the polynomial in the inner double sum gives $P(z) = (z - a) Q(z)$.

Skills to practice

Solving quadratics in $\mathbb{C}$

III.3 $n$-th roots of unity

The equation $z^n = 1$ has, in $\mathbb{R}$, one or two solutions ($1$ if $n$ is odd; $\pm 1$ if $n$ is even). In $\mathbb{C}$, it has exactly $n$ distinct solutions, and they are arranged in the simplest possible way: regularly spaced on the unit circle. These are the $n$-th roots of unity --- a finite set, denoted $\mathbb{U}_n$, stable under product and inverse (we will say later in the chapter on algebraic structures that $\mathbb{U}_n$ is a group), with a rich geometric image (a regular $n$-gon).

Definition — $n$-th roots of unity

For $n \in \mathbb{N}^*$, the set of $n$-th roots of unity is $$ \textcolor{colordef}{\mathbb{U}_n} = \{ z \in \mathbb{C} \ \mid \ z^n = 1 \}. $$

Example

A few small cases by hand: $\mathbb{U}_2 = \{1, -1\}$ (the real solutions of $z^2 = 1$); $\mathbb{U}_4 = \{1, i, -1, -i\}$. The general structure --- $n$ regularly-spaced points on the unit circle --- is the content of the next Theorem.

Theorem — Description of $\mathbb{U}_n$

For $n \in \mathbb{N}^*$ $$ \textcolor{colorprop}{\mathbb{U}_n = \left\{ e^{2ik\pi/n} \ \mid \ k \in \{0, 1, \dots, n-1\} \right\}}. $$ The set $\mathbb{U}_n$ has exactly $n$ elements; they sit on the unit circle, regularly spaced by an angle $2\pi/n$. Setting $\omega = e^{2i\pi/n}$, they read $1, \omega, \omega^2, \dots, \omega^{n-1}$.

Proof

Inclusion $\supset$. For $\omega_k = e^{2ik\pi/n}$, $\omega_k^n = e^{2ik\pi} = 1$, so $\omega_k \in \mathbb{U}_n$.
Inclusion $\subset$. If $z^n = 1$, then $|z|^n = 1$ in $\mathbb{R}_+$, so $|z| = 1$. Write $z = e^{i\theta}$. Then $z^n = e^{in\theta} = 1$ rewrites $n\theta \equiv 0 \,[2\pi]$, i.e. $\theta \equiv 0 \,[2\pi/n]$. So $\theta = 2k\pi/n$ for some $k \in \mathbb{Z}$, and $z = \omega_k$. Choosing $k$ modulo $n$ identifies $z$ with one of $\omega_0, \dots, \omega_{n-1}$.
Distinctness. The $\omega_k$ for $k = 0, \dots, n-1$ are distinct: $\omega_k = \omega_{k'}$ means $2k\pi/n \equiv 2k'\pi/n \,[2\pi]$, i.e. $k \equiv k' \,[n]$, false for distinct $k, k' \in \{0, \dots, n-1\}$.

Example

For small $n$, the $n$-th roots of unity sit on the unit circle as the vertices of a regular $n$-gon inscribed in $\mathbb{U}$, starting at $1$:

$\mathbb{U}_1 = \{ 1 \}$ $\quad$
$\mathbb{U}_2 = \{ 1, -1 \}$ $\quad$
$\mathbb{U}_3 = \{ 1, j, j^2 \}$ where $j = e^{2i\pi/3} = -\dfrac{1}{2} + i\dfrac{\sqrt 3}{2}$ and $j^2 = \conjugate{j} = -\dfrac{1}{2} - i\dfrac{\sqrt 3}{2}$ $\quad$
$\mathbb{U}_4 = \{ 1, i, -1, -i \}$ $\quad$
$\mathbb{U}_6 = \{ 1, e^{i\pi/3}, e^{2i\pi/3}, -1, e^{-2i\pi/3}, e^{-i\pi/3} \} $ $\quad$

Proposition — Sum and product properties of $\mathbb{U}_n$

For $n \ge 2$ and $\omega = e^{2i\pi/n}$:

$\textcolor{colorprop}{\sum_{k=0}^{n-1} \omega^k = 0}$ (sum of all $n$-th roots of unity is zero).
$\textcolor{colorprop}{\prod_{k=0}^{n-1} \omega^k = (-1)^{n-1}}$.
For all $z \in \mathbb{C}$: $\textcolor{colorprop}{z^n - 1 = \prod_{k=0}^{n-1} (z - \omega^k)}$.

Proof

Sum. Geometric series with ratio $\omega \ne 1$: $\sum_{k=0}^{n-1} \omega^k = (1 - \omega^n)/(1 - \omega) = (1 - 1)/(1 - \omega) = 0$.
Product. $\prod_{k=0}^{n-1} \omega^k = \omega^{0+1+\cdots+(n-1)} = \omega^{n(n-1)/2} = (e^{2i\pi/n})^{n(n-1)/2} = e^{i\pi(n-1)} = (-1)^{n-1}$.
Factorization. The polynomial $z^n - 1$ has leading coefficient $1$ and admits $1, \omega, \dots, \omega^{n-1}$ as $n$ distinct roots. Iterating the Factorization by a root Proposition above on these $n$ distinct roots — each step lowering the degree by one while keeping leading coefficient $1$ — gives $z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k)$ (the final quotient is the constant $1$, by a degree and leading-coefficient count).

Example

The cube roots of unity $\mathbb{U}_3 = \{1, j, j^2\}$ satisfy $1 + j + j^2 = 0$. In particular, $j^2 = -1 - j$, so any polynomial expression in $j$ reduces to $\alpha + \beta j$ with $\alpha, \beta \in \mathbb{R}$ --- this is the workhorse of cube-root computations. For instance, $j^3 = 1$, $j^4 = j$, $j^5 = j^2 = -1-j$, and $(1+j)(1+j^2) = 1 + j + j^2 + j^3 = 0 + 1 = 1$.

Skills to practice

Manipulating $n$-th roots of unity

III.4 $n$-th roots of any complex

Once we know the $n$-th roots of $1$, finding the $n$-th roots of any nonzero complex is one short step away. We pick any $n$-th root $\delta_0$ of the target $a$ (built from the trigonometric form), and then all $n$-th roots are obtained by multiplying $\delta_0$ by elements of $\mathbb{U}_n$. The resulting set is again a regular $n$-gon, this time centered at the origin with one vertex at $\delta_0$.

Theorem — $n$-th roots of a complex

Let $a \in \mathbb{C}^*$ and $n \in \mathbb{N}^*$. Write $a = \rho\, e^{i\theta}$. The equation $z^n = a$ has exactly $n$ solutions in $\mathbb{C}$: $$ \textcolor{colorprop}{z_k = \rho^{1/n}\, e^{i(\theta + 2k\pi)/n} \qquad k \in \{0, 1, \dots, n-1\}}. $$ Equivalently, if $\delta_0 = \rho^{1/n}\, e^{i\theta/n}$ is a particular root and $\omega = e^{2i\pi/n}$, then the $n$ solutions are $\delta_0, \delta_0 \omega, \delta_0 \omega^2, \dots, \delta_0 \omega^{n-1}$.

Proof

$z^n = a$ rewrites $(z/\delta_0)^n = a/\delta_0^n = 1$, i.e. $z/\delta_0 \in \mathbb{U}_n$. Hence $z = \delta_0 \omega^k$ for some $k \in \{0, \dots, n-1\}$. Each $k$ gives a distinct $z$ (multiplication by $\delta_0 \ne 0$ is injective), so there are exactly $n$ solutions.

Method — Compute the $n$-th roots of a complex

For $z^n = a$ with $a \ne 0$:

put $a$ in trigonometric form $\rho\, e^{i\theta}$;
write a particular root $\delta_0 = \rho^{1/n} e^{i\theta/n}$;
multiply by the $n$ elements of $\mathbb{U}_n$ to get all $n$ solutions.

Geometrically: place the first root, then rotate $n-1$ times by $2\pi/n$.

Example

Find the cube roots of $a = 8i$.

Answer

$|a| = 8$, $\arg(a) = \pi/2$, so $a = 8\, e^{i\pi/2}$. A particular cube root is $\delta_0 = 2\, e^{i\pi/6} = \sqrt 3 + i$. The three cube roots: $$ z_0 = \sqrt 3 + i \quad z_1 = 2\, e^{i(\pi/6 + 2\pi/3)} = 2\, e^{i\,5\pi/6} = -\sqrt 3 + i \quad z_2 = 2\, e^{i(\pi/6 + 4\pi/3)} = 2\, e^{-i\pi/2} = -2i. $$

Example

The three cube roots of $8i$ form an equilateral triangle on the circle of radius $\sqrt[3]{8} = 2$. (The point $a = 8i$ would be off the picture above; the diagram is centered on the roots.)

Skills to practice

Computing $n$-th roots of a complex

IV Geometric interpretation

We now cash in the bridge between $\mathbb{C}$ and the Euclidean plane. To each point $M$ of the plane we associate its affix $z_M \in \mathbb{C}$, with the dictionary: a vector becomes a difference of affixes, a translation an addition, a rotation a multiplication by $e^{i\theta}$. Alignment becomes « ratio is real »; orthogonality becomes « ratio is pure imaginary ». The maps $z \mapsto az + b$, called direct similitudes, encode the entire family of « rotation $\circ$ homothety $\circ$ translation ».

IV.1 Affix$\virgule$ alignment$\virgule$ orthogonality

The Argand identification is bidirectional: a complex $z = a + ib$ corresponds to the point $M(a, b)$, and a point $M$ has affix $z_M = a + ib$. Vectors live in this dictionary too: the vector $\overrightarrow{AB}$ has affix $z_B - z_A$, which encodes both the magnitude (its modulus $= |z_B - z_A| = AB$) and the direction (its argument). From this the geometric tests --- alignment, orthogonality, angle --- become routine algebra.

Definition — Affix

Fix an orthonormal direct frame $(O, \vec u, \vec v)$ of the plane.

The affix of the point $M$ of coordinates $(a, b)$ is $\textcolor{colordef}{z_M} = a + ib$.
The affix of the vector $\vec w$ of coordinates $(a, b)$ is $\textcolor{colordef}{z_{\vec w}} = a + ib$.
In particular, the affix of $\overrightarrow{AB}$ is $z_B - z_A$.

The map $M \mapsto z_M$ is a bijection from the plane onto $\mathbb{C}$.

Proposition — Distance$\virgule$ alignment$\virgule$ orthogonality

For three distinct points $A, B, C$ of the plane:

$\textcolor{colorprop}{AB = |z_B - z_A|}$ (distance is modulus of vector affix).
$A, B, C$ are aligned $\iff \textcolor{colorprop}{\dfrac{z_C - z_A}{z_B - z_A} \in \mathbb{R}}$.
$\overrightarrow{AB} \perp \overrightarrow{AC} \iff \textcolor{colorprop}{\dfrac{z_C - z_A}{z_B - z_A} \in i\mathbb{R}}$.
More generally, the oriented angle from $\overrightarrow{AB}$ to $\overrightarrow{AC}$ is $\textcolor{colorprop}{\arg\!\left( \dfrac{z_C - z_A}{z_B - z_A} \right) \,[2\pi]}$.

Proof

Distance. $z_B - z_A$ has modulus $\sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = AB$.
Alignment. $A, B, C$ are aligned (with $A \ne B$) iff $\overrightarrow{AC}$ and $\overrightarrow{AB}$ are colinear, i.e. there is $\lambda \in \mathbb{R}$ with $z_C - z_A = \lambda(z_B - z_A)$. Dividing by $z_B - z_A \ne 0$, this means $(z_C - z_A)/(z_B - z_A) = \lambda \in \mathbb{R}$.
Orthogonality. The dot product is $\operatorname{Re}\bigl( \conjugate{z_B - z_A}(z_C - z_A) \bigr)$. It vanishes iff $\conjugate{z_B - z_A}(z_C - z_A)$ is pure imaginary, iff $(z_C - z_A)/(z_B - z_A) = (z_C - z_A)\conjugate{z_B - z_A}/|z_B - z_A|^2$ is pure imaginary (multiplication by the positive real $1/|z_B - z_A|^2$ preserves the imaginary axis).
Angle. The argument of a quotient of affixes is the difference of the arguments of the corresponding vectors --- which is the oriented angle.

Method — Prove alignment or orthogonality

Given three points $A, B, C$ by their affixes:

to test alignment, compute the ratio $r = (z_C - z_A)/(z_B - z_A)$ and check whether $r \in \mathbb{R}$ (e.g. by verifying $r = \conjugate{r}$, or by computing $\operatorname{Im}(r)$);
to test orthogonality, check whether $r \in i\mathbb{R}$ (e.g. $r = -\conjugate{r}$, or $\operatorname{Re}(r) = 0$);
to compute the angle, take $\arg(r)$.

This converts geometric questions into one-line algebraic verifications.

Example

Are $A(1+i)$, $B(2+3i)$, $C(-1-3i)$ aligned?
$r = (z_C - z_A)/(z_B - z_A) = (-2 - 4i)/(1 + 2i)$. Multiplying by $\conjugate{1 + 2i} = 1 - 2i$: $$ r = \frac{(-2 - 4i)(1 - 2i)}{|1+2i|^2} = \frac{-2 + 4i - 4i + 8i^2}{5} = \frac{-2 - 8}{5} = -2. $$ $r \in \mathbb{R}$, so $A, B, C$ are aligned.

Example

Show that the triangle $A(0)$, $B(1)$, $C(\dfrac{1}{2} + \dfrac{i\sqrt 3}{2})$ is equilateral.

Answer

$AB = |1 - 0| = 1$. $AC = |\dfrac{1}{2} + \dfrac{i\sqrt 3}{2}| = \sqrt{1/4 + 3/4} = 1$. $BC = |\dfrac{1}{2} + \dfrac{i\sqrt 3}{2} - 1| = |\!-\!\dfrac{1}{2} + \dfrac{i\sqrt 3}{2}| = 1$. Three equal sides, equilateral.
Alternative: $C - A = e^{i\pi/3}(B - A)$, so $\overrightarrow{AC}$ is obtained from $\overrightarrow{AB}$ by rotation of angle $\pi/3$ --- the signature of an equilateral triangle.

Skills to practice

Testing alignment$\virgule$ orthogonality$\virgule$ and angles

IV.2 Direct similitudes

The general direct similitude of the plane has the form $z \mapsto az + b$ with $a \in \mathbb{C}^*$ and $b \in \mathbb{C}$. The complex number $a$ encodes a rotation (by argument) and a scaling (by modulus); the constant $b$ encodes a translation. Together, four classical transformations of the plane fit in this single algebraic mold: translations, homotheties, rotations, and the general similitude. We disentangle them here.

Definition — Direct similitude

A direct similitude of the plane is a map of the form $$ \textcolor{colordef}{f: z \mapsto az + b} \qquad a \in \mathbb{C}^*, \ b \in \mathbb{C}. $$ The number $|a|$ is called the ratio, and $\arg(a) \,[2\pi]$ the angle of $f$.

Example

$f: z \mapsto 2z$ is a direct similitude with ratio $|2| = 2$ and angle $\arg(2) = 0$ (a homothety).
$f: z \mapsto i z$ is a direct similitude with ratio $1$ and angle $\pi/2$ (a quarter-turn rotation around $0$).
$f: z \mapsto z + 1$ is a direct similitude with ratio $1$ and angle $0$ (a translation).
$f: z \mapsto (1 + i)z + 2$ is a direct similitude with ratio $\sqrt 2$ and angle $\pi/4$.

Proposition — Fixed point and structure

Let $f: z \mapsto az + b$ be a direct similitude.

If $a = 1$, $f$ is the translation of vector of affix $b$ ($f$ has no fixed point unless $b = 0$).
If $a \ne 1$, $f$ has a unique fixed point $\textcolor{colorprop}{\omega = \dfrac{b}{1 - a}}$, and $f$ writes $$ \textcolor{colorprop}{f(z) = \omega + a(z - \omega)}. $$ The point $\omega$ is called the center of $f$.

Proof

Case $a = 1$. $f(z) = z + b$, so $f(z) = z \iff b = 0$.
Case $a \ne 1$. $f(z) = z \iff (1-a)z = b \iff z = b/(1-a) =: \omega$. Then $f(z) - \omega = az + b - \omega = a(z - \omega) + (a\omega + b - \omega) = a(z - \omega) + (b - (1-a)\omega) = a(z - \omega) + 0$, hence $f(z) = \omega + a(z - \omega)$.

Method — Classify a direct similitude

Given $f: z \mapsto az + b$:

if $a = 1$: translation of vector $\vec b$ (affix $b$);
if $a \in \mathbb{R}^*\setminus\{1\}$: homothety of center $\omega = b/(1-a)$, ratio $a$;
if $|a| = 1$ and $a \ne 1$: rotation of center $\omega = b/(1-a)$, angle $\arg(a)$;
otherwise (general $a$): direct similitude of center $\omega = b/(1-a)$, ratio $|a|$, angle $\arg(a)$. It is the composition (in any order, around $\omega$) of the homothety of ratio $|a|$ and the rotation of angle $\arg(a)$.

Example

Identify $f: z \mapsto (1 + i) z + 2$.

Answer

$a = 1 + i = \sqrt 2\, e^{i\pi/4}$ ($|a| = \sqrt 2$, $\arg(a) = \pi/4$). Fixed point: $\omega = 2/(1 - 1 - i) = 2/(-i) = 2i$. Hence $f$ is the direct similitude of center $2i$, ratio $\sqrt 2$, angle $\pi/4$.

Example

A direct similitude maps a triangle to a similar triangle (same shape, possibly different size and orientation), with all vertices rotated and scaled around the center $\omega$.

Skills to practice

Classifying $z \mapsto az + b$

IV.3 Special cases: translations$\virgule$ homotheties$\virgule$ rotations

We unpack the three classical transformations as special direct similitudes. Each has a one-line algebraic form, a one-line geometric description, and a single parameter to identify (vector, ratio, or angle).

Definition — Translation$\virgule$ homothety$\virgule$ rotation

The translation of vector $\vec u$ of affix $b \in \mathbb{C}$ is $\textcolor{colordef}{t_{\vec u}: z \mapsto z + b}$.
The homothety of center $\omega \in \mathbb{C}$ and ratio $k \in \mathbb{R}^*$ is $\textcolor{colordef}{h_{\omega, k}: z \mapsto \omega + k(z - \omega)} = kz + (1-k)\omega$.
The rotation of center $\omega \in \mathbb{C}$ and angle $\theta \in \mathbb{R}$ is $\textcolor{colordef}{r_{\omega, \theta}: z \mapsto \omega + e^{i\theta}(z - \omega)} = e^{i\theta}z + (1 - e^{i\theta})\omega$.

Example

Translation $z \mapsto z + b$: every point shifts by the same vector $\vec b$.

Example

Homothety $h_{\omega, 2}$: each $z$ is sent twice as far from $\omega$, on the same ray. Negative ratio inverts the direction (point symmetry when $k = -1$).

Example

Rotation $r_{\omega, \theta}$: each point turns around $\omega$ by angle $\theta$, keeping its distance to $\omega$ unchanged.

Proposition — Composition of similitudes

Let $f: z \mapsto a_1 z + b_1$ and $g: z \mapsto a_2 z + b_2$ be two direct similitudes. Then $$ \textcolor{colorprop}{(f \circ g)(z) = a_1 a_2\, z + a_1 b_2 + b_1}. $$ In particular, the composition of two direct similitudes is a direct similitude, with $a$-coefficient $a_1 a_2$.

Proof

$(f \circ g)(z) = f(g(z)) = f(a_2 z + b_2) = a_1(a_2 z + b_2) + b_1 = a_1 a_2 z + a_1 b_2 + b_1$.

Example

Compose two rotations: $r_1 = r_{0, \pi/3}: z \mapsto e^{i\pi/3} z$ and $r_2 = r_{0, \pi/4}: z \mapsto e^{i\pi/4} z$. Then $r_2 \circ r_1: z \mapsto e^{i\pi/4} e^{i\pi/3} z = e^{i\,7\pi/12} z$, which is the rotation of center $0$ and angle $7\pi/12$. Composition of two rotations of same center adds angles --- a fact made obvious by the algebraic form.

Skills to practice

Composing classical transformations