Complex Numbers

Complex Numbers

Complex numbers extend the real number system by introducing a solution to the equation $x^2 + 1 = 0$. This seemingly abstract idea turns out to be indispensable in pure mathematics, engineering, and physics, providing powerful tools for analysing rotations, oscillations, and polynomial equations.

Board Coverage

Board	Paper	Notes
AQA	Paper 1	Full coverage including transformations $w = f(z)$
Edexcel	FP1/FP2	De Moivre, roots of unity, loci in FP2
OCR (A)	Paper 1	De Moivre, exponential form, roots of unity
CIE	P1	Loci required; exponential form and roots of unity in depth

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1. Review of A Level Complex Numbers

Definition. A complex number is a number of the form $z = a + bi$ where $a, b \in \mathbb{R}$ and $i$ is defined by the property $i^2 = -1$. The set of all complex numbers is denoted $\mathbb{C}$.

We call $a$ the real part (written $\operatorname{Re}(z)$) and $b$ the imaginary part (written $\operatorname{Im}(z)$). Two complex numbers are equal if and only if both their real and imaginary parts are equal.

1.1 The Argand Diagram

Definition. The Argand diagram is a representation of $\mathbb{C}$ as a plane, where the horizontal axis represents $\operatorname{Re}(z)$ and the vertical axis represents $\operatorname{Im}(z)$. The complex number $z = a + bi$ is plotted as the point $(a, b)$.

1.2 Modulus and Argument

Definition. For $z = a + bi \neq 0$:

• The modulus $|z|$ is defined as $|z| = \sqrt{a^2 + b^2}$.
• The argument $\arg(z)$ is the angle $\theta$ (measured anticlockwise from the positive real axis) such that $\tan\theta = \dfrac{b}{a}$, with $-\pi < \theta \leq \pi$ (the principal argument).

$\boxed{z = a + bi = |z|(\cos\theta + i\sin\theta) = r(\cos\theta + i\sin\theta)}$

where $r = |z|$ and $\theta = \arg(z)$.

1.3 Arithmetic with Complex Numbers

For $z_1 = a + bi$ and $z_2 = c + di$:

$$\begin{aligned} z_1 + z_2 &= (a + c) + (b + d)i \\ z_1 z_2 &= (ac - bd) + (ad + bc)i \\ \frac{z_1}{z_2} &= \frac◆LB◆z_1 \overline{z_2}◆RB◆◆LB◆|z_2|^2◆RB◆ = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} \end{aligned}$$

where $\overline{z_2} = c - di$ is the complex conjugate of $z_2$.

warning $\theta = \arctan(b/a)$ only gives the correct principal argument when $a > 0$.

Worked Example: Modulus, argument, and polar form

Find the modulus, argument, and polar form of $z = 1 - \sqrt{3}\,i$.

$|z| = \sqrt◆LB◆1^2 + (-\sqrt{3})^2◆RB◆ = \sqrt{1 + 3} = 2$

Since $(a, b) = (1, -\sqrt{3})$ lies in the fourth quadrant:

$\arg(z) = \arctan\!\left(\frac◆LB◆-\sqrt{3}◆RB◆◆LB◆1◆RB◆\right) = -\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$

Polar form: $z = 2\!\left(\cos\!\left(-\dfrac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\right) + i\sin\!\left(-\dfrac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\right)\right)$.

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2. De Moivre's Theorem

Theorem (De Moivre). For any integer $n$ and any angle $\theta$:

$\boxed{\left(\cos\theta + i\sin\theta\right)^n = \cos(n\theta) + i\sin(n\theta)}$

Proof of De Moivre's Theorem (by induction for $n \geq 0$)

Base case ($n = 0$): $(\cos\theta + i\sin\theta)^0 = 1 = \cos 0 + i\sin 0$. ✓

Inductive step. Assume $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$ for some $k \geq 0$. Then:

$$\begin{aligned} (\cos\theta + i\sin\theta)^{k+1} &= (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta) \\ &= [\cos(k\theta) + i\sin(k\theta)] \cdot [\cos\theta + i\sin\theta] \\ &= \cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta + i\,[\cos(k\theta)\sin\theta + \sin(k\theta)\cos\theta] \\ &= \cos((k+1)\theta) + i\sin((k+1)\theta) \end{aligned}$$

using the compound angle identities. ✓

For negative integers, note that $\dfrac◆LB◆1◆RB◆◆LB◆\cos\theta + i\sin\theta◆RB◆ = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$, so the result follows. $\square$

Intuition. De Moivre's theorem says that raising a complex number on the unit circle to the $n$-th power simply multiplies its angle by $n$. This is because multiplication of complex numbers adds their arguments: $r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)}$.

2.1 Applications: Trigonometric Identities

De Moivre's theorem provides a systematic way to derive multiple-angle identities.

Example. Express $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By De Moivre: $\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3$.

Expanding the RHS using the binomial theorem:

$$\begin{aligned} (\cos\theta + i\sin\theta)^3 &= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta \end{aligned}$$

Equating real and imaginary parts:

$\boxed{\cos 3\theta = 4\cos^3\theta - 3\cos\theta}$

$\boxed{\sin 3\theta = 3\sin\theta - 4\sin^3\theta}$

2.2 Powers of Complex Numbers

To compute $z^n$ where $z = r(\cos\theta + i\sin\theta)$:

$z^n = r^n\left(\cos(n\theta) + i\sin(n\theta)\right)$

Worked Example: Computing a high power

Find $(1 + i)^{10}$.

First write in polar form: $1 + i = \sqrt{2}\!\left(\cos\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ + i\sin\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\right)$.

$$\begin{aligned} (1 + i)^{10} &= \left(\sqrt{2}\right)^{10}\!\left(\cos\frac◆LB◆10\pi◆RB◆◆LB◆4◆RB◆ + i\sin\frac◆LB◆10\pi◆RB◆◆LB◆4◆RB◆\right) \\ &= 32\!\left(\cos\frac◆LB◆5\pi◆RB◆◆LB◆2◆RB◆ + i\sin\frac◆LB◆5\pi◆RB◆◆LB◆2◆RB◆\right) \\ &= 32\!\left(\cos\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + i\sin\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right) \\ &= 32(0 + i) = 32i \end{aligned}$$

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3. Roots of Unity

Definition. The $n$-th roots of unity are the solutions to the equation $z^n = 1$ for $n \in \mathbb{Z}^+$.

By De Moivre's theorem, writing $1 = \cos 0 + i\sin 0 = \cos(2k\pi) + i\sin(2k\pi)$ for any integer $k$, the $n$ distinct solutions are:

$\boxed{z_k = \cos\!\left(\frac◆LB◆2k\pi◆RB◆◆LB◆n◆RB◆\right) + i\sin\!\left(\frac◆LB◆2k\pi◆RB◆◆LB◆n◆RB◆\right), \quad k = 0, 1, 2, \ldots, n-1}$

3.1 Geometric Interpretation

The $n$-th roots of unity lie on the unit circle $|z| = 1$ in the Argand diagram, equally spaced at angles of $\dfrac◆LB◆2\pi◆RB◆◆LB◆n◆RB◆$ radians apart. They form the vertices of a regular $n$-gon inscribed in the unit circle, with one vertex at $z = 1$.

3.2 Sum and Product of Roots

Since the roots satisfy $z^n - 1 = 0$, the sum of all $n$-th roots of unity is zero:

$\sum_{k=0}^{n-1} z_k = 0$

This follows from the coefficient of $z^{n-1}$ in $z^n - 1 = 0$ being zero (by Vieta's formulas). Equivalently, the roots form a geometric series with ratio $\omega = e^{2\pi i/n}$ and first term 1, giving:

$\sum_{k=0}^{n-1} \omega^k = \frac◆LB◆1 - \omega^n◆RB◆◆LB◆1 - \omega◆RB◆ = \frac◆LB◆1 - 1◆RB◆◆LB◆1 - \omega◆RB◆ = 0$

The product of all $n$-th roots of unity is:

$\prod_{k=0}^{n-1} z_k = (-1)^{n-1}$

Worked Example: Cube roots of unity

Find all cube roots of unity and verify that their sum is zero.

$z^3 = 1 \implies z_k = \cos\!\left(\dfrac◆LB◆2k\pi◆RB◆◆LB◆3◆RB◆\right) + i\sin\!\left(\dfrac◆LB◆2k\pi◆RB◆◆LB◆3◆RB◆\right)$ for $k = 0, 1, 2$.

$$\begin{aligned} z_0 &= \cos 0 + i\sin 0 = 1 \\ z_1 &= \cos\frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆ + i\sin\frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆ = -\frac{1}{2} + \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\,i \\ z_2 &= \cos\frac◆LB◆4\pi◆RB◆◆LB◆3◆RB◆ + i\sin\frac◆LB◆4\pi◆RB◆◆LB◆3◆RB◆ = -\frac{1}{2} - \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\,i \end{aligned}$$

Sum: $1 - \dfrac{1}{2} + \dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\,i - \dfrac{1}{2} - \dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\,i = 0$. ✓

Product: $1 \cdot \left(-\dfrac{1}{2} + \dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\,i\right)\left(-\dfrac{1}{2} - \dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆\,i\right) = 1 \cdot \left(\dfrac{1}{4} + \dfrac{3}{4}\right) = 1 = (-1)^{3-1}$. ✓

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4. Exponential Form

Definition. The exponential form of a complex number $z = r(\cos\theta + i\sin\theta)$ is:

$\boxed{z = re^{i\theta}}$

where $e^{i\theta} \equiv \cos\theta + i\sin\theta$ by Euler's formula.

Proof of Euler's Formula (from Maclaurin series)

The Maclaurin series for $e^x$, $\cos x$, and $\sin x$ are:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$

Substituting $x = i\theta$ into the series for $e^x$:

$$\begin{aligned} e^{i\theta} &= 1 + i\theta + \frac◆LB◆(i\theta)^2◆RB◆◆LB◆2!◆RB◆ + \frac◆LB◆(i\theta)^3◆RB◆◆LB◆3!◆RB◆ + \frac◆LB◆(i\theta)^4◆RB◆◆LB◆4!◆RB◆ + \frac◆LB◆(i\theta)^5◆RB◆◆LB◆5!◆RB◆ + \cdots \\ &= 1 + i\theta + \frac◆LB◆i^2\theta^2◆RB◆◆LB◆2!◆RB◆ + \frac◆LB◆i^3\theta^3◆RB◆◆LB◆3!◆RB◆ + \frac◆LB◆i^4\theta^4◆RB◆◆LB◆4!◆RB◆ + \frac◆LB◆i^5\theta^5◆RB◆◆LB◆5!◆RB◆ + \cdots \\ &= 1 + i\theta - \frac◆LB◆\theta^2◆RB◆◆LB◆2!◆RB◆ - \frac◆LB◆i\theta^3◆RB◆◆LB◆3!◆RB◆ + \frac◆LB◆\theta^4◆RB◆◆LB◆4!◆RB◆ + \frac◆LB◆i\theta^5◆RB◆◆LB◆5!◆RB◆ - \cdots \end{aligned}$$

using $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, $i^5 = i$, and so on. Grouping real and imaginary parts:

$e^{i\theta} = \underbrace◆LB◆\left(1 - \frac◆LB◆\theta^2◆RB◆◆LB◆2!◆RB◆ + \frac◆LB◆\theta^4◆RB◆◆LB◆4!◆RB◆ - \cdots\right)◆RB◆_{=\,\cos\theta} + i\underbrace◆LB◆\left(\theta - \frac◆LB◆\theta^3◆RB◆◆LB◆3!◆RB◆ + \frac◆LB◆\theta^5◆RB◆◆LB◆5!◆RB◆ - \cdots\right)◆RB◆_{=\,\sin\theta}$

Therefore $e^{i\theta} = \cos\theta + i\sin\theta$. $\square$

4.1 Euler's Identity

Setting $\theta = \pi$:

$\boxed{e^{i\pi} + 1 = 0}$

This celebrated identity connects five fundamental constants: $e$, $i$, $\pi$, $1$, and $0$.

Intuition. Euler's identity says that starting at the point $1$ on the real axis and rotating by $\pi$ radians (half a turn) on the unit circle lands you at $-1$. The exponential $e^{i\theta}$ describes a point moving around the unit circle at a rate determined by $\theta$.

4.2 Exponential Rules for Complex Numbers

The standard laws of indices extend naturally:

$$\begin{aligned} z_1 z_2 &= r_1 r_2 \, e^{i(\theta_1 + \theta_2)} & &\mathrm{(arguments add)} \\ \frac{z_1}{z_2} &= \frac{r_1}{r_2} \, e^{i(\theta_1 - \theta_2)} & &\mathrm{(arguments subtract)} \\ z^n &= r^n e^{in\theta} & &\mathrm{(argument multiplies)} \end{aligned}$$

info with the $\cos\theta + i\sin\theta$ form. All boards require De Moivre's theorem. :::

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5. Loci in the Argand Diagram

5.1 Circles: $|z - a| = r$

Definition. The locus $|z - a| = r$, where $a \in \mathbb{C}$ and $r \in \mathbb{R}^+$, is a circle with centre $a$ and radius $r$ in the Argand diagram.

$|z - a| = r \iff \sqrt◆LB◆(x - \alpha)^2 + (y - \beta)^2◆RB◆ = r \iff (x - \alpha)^2 + (y - \beta)^2 = r^2$

where $a = \alpha + \beta i$ and $z = x + yi$.

The region $|z - a| < r$ is the interior of the circle, and $|z - a| > r$ is the exterior.

5.2 Perpendicular Bisectors: $|z - a| = |z - b|$

This locus represents all points equidistant from $a$ and $b$, which is the perpendicular bisector of the line segment joining $a$ and $b$.

5.3 Half-Lines: $\arg(z - a) = \alpha$

Definition. The locus $\arg(z - a) = \alpha$ is a half-line (ray) starting from the point $a$ (not including $a$ itself) making an angle $\alpha$ with the positive real direction.

The region $\alpha_1 < \arg(z - a) < \alpha_2$ is an angular sector (wedge) with vertex at $a$.

5.4 Combined Loci and Regions

Exam questions often require describing a region defined by combining loci, such as:

• $|z - 3| \leq 2$ and $\arg(z) \geq \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$: the part of a disc in a sector.

A half-line $\arg(z - a) = \alpha$ does not include the point $a$. When shading

regions, be careful about whether boundaries are included (solid line) or excluded (dashed line).

Always sketch loci problems. The algebraic description follows from the geometric picture.

Common exam technique: identify the boundary (circle, line, half-line), then determine which side of the boundary is included by testing a point. :::

Worked Example: Describing a locus algebraically

A complex number $z$ satisfies $|z - 2i| \leq 3$ and $0 \leq \arg(z) \leq \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$. Find the greatest possible value of $|z|$ and the least possible value of $|z|$.

The first condition: $|z - 2i| \leq 3$ is the closed disc of radius 3 centred at $2i$, i.e. at $(0, 2)$.

The second condition: $0 \leq \arg(z) \leq \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$ restricts $z$ to the first quadrant (including axes).

The disc centre $(0, 2)$ with radius 3 extends from $y = -1$ to $y = 5$ and from $x = -3$ to $x = 3$.

Greatest $|z|$: The point in the region farthest from the origin is where the boundary of the disc intersects the first quadrant boundary furthest from the origin. The disc intersects the positive $y$-axis at $(0, 5)$, giving $|z| = 5$.

Least $|z|$: We need the closest point in the region to the origin. The disc boundary is $(x^2 + (y-2)^2) = 9$. The closest point on this circle to the origin lies along the line from the origin through the centre $(0,2)$, which is the $y$-axis. The point $(0, -1)$ is outside the first quadrant. Within the first quadrant, the closest point is where the circle meets the $x$-axis: setting $y = 0$, $x^2 + 4 = 9 \implies x = \sqrt{5}$. So $|z| = \sqrt{5}$.

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6. Complex Transformations

6.1 The Mapping $w = f(z)$

A complex transformation is a function $w = f(z)$ that maps points in the $z$-plane (Argand diagram for $z$) to points in the $w$-plane (Argand diagram for $w$).

6.2 Linear Transformations: $w = az + b$

For $w = az + b$ where $a, b \in \mathbb{C}$ and $a \neq 0$:

Writing $a = \lambda e^{i\alpha}$ and $b = \mu e^{i\beta}$:

1. $|a| = \lambda$ produces an enlargement (scale factor $\lambda$) about the origin.
2. $\arg(a) = \alpha$ produces a rotation through angle $\alpha$ about the origin.
3. $b$ produces a translation by the vector representing $b$.

The composition is: enlarge by $|a|$, rotate by $\arg(a)$, then translate by $b$.

6.3 Inversion: $w = \dfrac{1}{z}$

The transformation $w = \dfrac{1}{z}$ maps:

• Circles not through the origin to circles.
• Circles through the origin to straight lines not through the origin.
• Straight lines through the origin to straight lines through the origin.
• Straight lines not through the origin to circles through the origin.

6.4 Reciprocal: $w = z + \dfrac{1}{z}$ and $w = \dfrac{z - 1}{z + 1}$

These are common in exam questions. The general approach is:

1. Express $z$ in terms of $w$: $z = f^{-1}(w)$.
2. Apply the given condition on $z$ (e.g. $|z| = 2$) to find the locus of $w$.

Critical points. A critical point of a transformation $w = f(z)$ is a point $z_0$ where $f'(z_0) = 0$. At a critical point, the mapping is not conformal (angles are not preserved).

Worked Example: Image of a line under inversion

Find the image of the line $\operatorname{Re}(z) = 1$ under the transformation $w = \dfrac{1}{z}$.

Let $z = x + yi$ with $x = 1$, so $z = 1 + yi$ and $y \in \mathbb{R}$.

$w = \frac{1}{1 + yi} = \frac{1 - yi}{1 + y^2} = \frac{1}{1 + y^2} - \frac{y}{1 + y^2}\,i$

Let $w = u + vi$. Then $u = \dfrac{1}{1 + y^2}$ and $v = \dfrac{-y}{1 + y^2}$.

Note that $v = -uy$, so $y = -\dfrac{v}{u}$ (when $u \neq 0$).

Substituting: $u = \dfrac{1}{1 + v^2/u^2} = \dfrac{u^2}{u^2 + v^2}$, giving $u^2 + v^2 = u$, i.e.:

$u^2 - u + v^2 = 0 \implies \left(u - \frac{1}{2}\right)^2 + v^2 = \frac{1}{4}$

This is a circle with centre $\left(\dfrac{1}{2}, 0\right)$ and radius $\dfrac{1}{2}$ in the $w$-plane.

AQA places significant emphasis on complex transformations including $w = f(z)$ mappings.

Edexcel and OCR cover this topic with less depth. CIE focuses more on loci than on transformations.

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7. Summary of Key Results

$\boxed{(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)}$

$\boxed{e^{i\theta} = \cos\theta + i\sin\theta}$

$\boxed{e^{i\pi} + 1 = 0}$

$\boxed{z_k = e^{2k\pi i/n} = \cos\!\left(\frac◆LB◆2k\pi◆RB◆◆LB◆n◆RB◆\right) + i\sin\!\left(\frac◆LB◆2k\pi◆RB◆◆LB◆n◆RB◆\right), \quad k = 0, 1, \ldots, n-1}$

$\boxed{\sum_{k=0}^{n-1} z_k = 0}$

$\boxed{|z - a| = r \iff \mathrm{circle centre } a \mathrm{ radius } r}$

$\boxed{\arg(z - a) = \alpha \iff \mathrm{half-line from } a \mathrm{ at angle } \alpha}$

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Problems

Problem 1. Express $z = -\sqrt{3} + i$ in modulus-argument form and hence find $z^5$ in the form $a + bi$.

Hint

Find $|z|$ and $\arg(z)$ first. Then apply De Moivre's theorem.

Answer

$|z| = \sqrt{3 + 1} = 2$.

The point $(-\sqrt{3}, 1)$ is in the second quadrant. $\arg(z) = \pi - \arctan\!\left(\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{3}◆RB◆\right) = \pi - \dfrac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ = \dfrac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆$.

$z^5 = 2^5\!\left(\cos\frac◆LB◆25\pi◆RB◆◆LB◆6◆RB◆ + i\sin\frac◆LB◆25\pi◆RB◆◆LB◆6◆RB◆\right) = 32\!\left(\cos\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ + i\sin\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆\right) = 32\!\left(\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ + \frac{1}{2}\,i\right) = 16\sqrt{3} + 16i$

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Problem 2. Use De Moivre's theorem to prove that $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$.

Hint

Expand $(\cos\theta + i\sin\theta)^4$ using the binomial theorem and equate real parts.

Answer

$\cos 4\theta + i\sin 4\theta = (\cos\theta + i\sin\theta)^4$.

Expanding: $\cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta$.

Real part: $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$.

Using $\sin^2\theta = 1 - \cos^2\theta$:

$$\begin{aligned} \cos 4\theta &= \cos^4\theta - 6\cos^2\theta(1 - \cos^2\theta) + (1 - \cos^2\theta)^2 \\ &= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + 1 - 2\cos^2\theta + \cos^4\theta \\ &= 8\cos^4\theta - 8\cos^2\theta + 1 \quad \square \end{aligned}$$

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Problem 3. Find all solutions to $z^4 = 16i$, expressing each in the form $a + bi$.

Hint

Write $16i = 16e^{i\pi/2}$ and use the roots formula.

Answer

$16i = 16\!\left(\cos\dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + i\sin\dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right)$.

$z_k = 2\!\left(\cos\!\left(\frac◆LB◆\pi/2 + 2k\pi◆RB◆◆LB◆4◆RB◆\right) + i\sin\!\left(\frac◆LB◆\pi/2 + 2k\pi◆RB◆◆LB◆4◆RB◆\right)\right), \quad k = 0, 1, 2, 3$

$$\begin{aligned} k = 0:\quad z_0 &= 2\!\left(\cos\frac◆LB◆\pi◆RB◆◆LB◆8◆RB◆ + i\sin\frac◆LB◆\pi◆RB◆◆LB◆8◆RB◆\right) = 2\!\left(\frac◆LB◆\sqrt◆LB◆2+\sqrt{2}◆RB◆◆RB◆◆LB◆2◆RB◆ + i\,\frac◆LB◆\sqrt◆LB◆2-\sqrt{2}◆RB◆◆RB◆◆LB◆2◆RB◆\right) = \sqrt◆LB◆2+\sqrt{2}◆RB◆ + i\sqrt◆LB◆2-\sqrt{2}◆RB◆ \\ k = 1:\quad z_1 &= 2\!\left(\cos\frac◆LB◆5\pi◆RB◆◆LB◆8◆RB◆ + i\sin\frac◆LB◆5\pi◆RB◆◆LB◆8◆RB◆\right) = -\sqrt◆LB◆2-\sqrt{2}◆RB◆ + i\sqrt◆LB◆2+\sqrt{2}◆RB◆ \\ k = 2:\quad z_2 &= 2\!\left(\cos\frac◆LB◆9\pi◆RB◆◆LB◆8◆RB◆ + i\sin\frac◆LB◆9\pi◆RB◆◆LB◆8◆RB◆\right) = -\sqrt◆LB◆2+\sqrt{2}◆RB◆ - i\sqrt◆LB◆2-\sqrt{2}◆RB◆ \\ k = 3:\quad z_3 &= 2\!\left(\cos\frac◆LB◆13\pi◆RB◆◆LB◆8◆RB◆ + i\sin\frac◆LB◆13\pi◆RB◆◆LB◆8◆RB◆\right) = \sqrt◆LB◆2-\sqrt{2}◆RB◆ - i\sqrt◆LB◆2+\sqrt{2}◆RB◆ \end{aligned}$$

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Problem 4. The fifth roots of unity are $\omega^0, \omega^1, \omega^2, \omega^3, \omega^4$ where $\omega = e^{2\pi i/5}$. Show that $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ and deduce that $\cos\dfrac◆LB◆2\pi◆RB◆◆LB◆5◆RB◆ + \cos\dfrac◆LB◆4\pi◆RB◆◆LB◆5◆RB◆ = -\dfrac{1}{2}$.

Hint

Sum the geometric series. Then separate real and imaginary parts.

Answer

The roots satisfy $z^5 - 1 = 0$. The coefficient of $z^4$ is 0, so by Vieta's formulas, $\sum_{k=0}^{4} \omega^k = 0$.

Alternatively: $\displaystyle\sum_{k=0}^{4}\omega^k = \frac◆LB◆1 - \omega^5◆RB◆◆LB◆1 - \omega◆RB◆ = \frac◆LB◆1 - 1◆RB◆◆LB◆1 - \omega◆RB◆ = 0$.

Expanding using $\omega^k = \cos\frac◆LB◆2k\pi◆RB◆◆LB◆5◆RB◆ + i\sin\frac◆LB◆2k\pi◆RB◆◆LB◆5◆RB◆$:

$\sum_{k=0}^{4}\omega^k = \underbrace◆LB◆\sum_{k=0}^{4}\cos\frac◆LB◆2k\pi◆RB◆◆LB◆5◆RB◆◆RB◆_{\mathrm{real}} + i\underbrace◆LB◆\sum_{k=0}^{4}\sin\frac◆LB◆2k\pi◆RB◆◆LB◆5◆RB◆◆RB◆_{\mathrm{imaginary}} = 0$

The imaginary part is zero by symmetry ($\sin\theta = -\sin(2\pi - \theta)$). The real part gives:

$1 + \cos\frac◆LB◆2\pi◆RB◆◆LB◆5◆RB◆ + \cos\frac◆LB◆4\pi◆RB◆◆LB◆5◆RB◆ + \cos\frac◆LB◆6\pi◆RB◆◆LB◆5◆RB◆ + \cos\frac◆LB◆8\pi◆RB◆◆LB◆5◆RB◆ = 0$

Since $\cos\frac◆LB◆6\pi◆RB◆◆LB◆5◆RB◆ = \cos\frac◆LB◆4\pi◆RB◆◆LB◆5◆RB◆$ and $\cos\frac◆LB◆8\pi◆RB◆◆LB◆5◆RB◆ = \cos\frac◆LB◆2\pi◆RB◆◆LB◆5◆RB◆$:

$1 + 2\cos\frac◆LB◆2\pi◆RB◆◆LB◆5◆RB◆ + 2\cos\frac◆LB◆4\pi◆RB◆◆LB◆5◆RB◆ = 0 \implies \cos\frac◆LB◆2\pi◆RB◆◆LB◆5◆RB◆ + \cos\frac◆LB◆4\pi◆RB◆◆LB◆5◆RB◆ = -\frac{1}{2} \quad \square$

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Problem 5. Sketch on separate Argand diagrams the loci given by (a) $|z - 1 - i| = |z - 3 + i|$, and (b) $\arg(z - 2) = \dfrac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$. Find the complex number(s) satisfying both conditions simultaneously.

Hint

Part (a) is a perpendicular bisector. Part (b) is a half-line. Find their intersection.

Answer

(a) $|z - (1 + i)| = |z - (3 - i)|$ is the perpendicular bisector of the segment joining $(1, 1)$ and $(3, -1)$. The midpoint is $(2, 0)$ and the slope of the segment is $\dfrac{-1 - 1}{3 - 1} = -1$, so the perpendicular bisector has slope $1$ and equation $y = x - 2$.

(b) $\arg(z - 2) = \dfrac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$ is a half-line from $(2, 0)$ at angle $\dfrac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$ to the positive real axis. Its equation is $y = \sqrt{3}(x - 2)$ for $x > 2$.

Intersection: Setting $x - 2 = \sqrt{3}(x - 2)$:

$(x - 2)(1 - \sqrt{3}) = 0$, so $x = 2$ (gives $y = 0$, but the half-line requires $x > 2$) or $1 = \sqrt{3}$, which is false.

There is no intersection. The half-line from $(2, 0)$ at angle $\pi/3$ has slope $\sqrt{3}$, while the perpendicular bisector has slope 1, and they only meet at the point $(2, 0)$ which is excluded from the half-line.

Answer: No complex number satisfies both conditions simultaneously.

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Problem 6. Find the image of the circle $|z| = 2$ under the transformation $w = \dfrac{z + 1}{z - 1}$.

Hint

Express $z$ in terms of $w$ and substitute $|z| = 2$.

Answer

$w = \dfrac{z + 1}{z - 1} \implies wz - w = z + 1 \implies z(w - 1) = w + 1 \implies z = \dfrac{w + 1}{w - 1}$.

Since $|z| = 2$:

$\left|\frac{w + 1}{w - 1}\right| = 2 \implies |w + 1| = 2|w - 1|$

Let $w = u + vi$:

$\sqrt{(u+1)^2 + v^2} = 2\sqrt{(u-1)^2 + v^2}$

Squaring: $(u+1)^2 + v^2 = 4[(u-1)^2 + v^2]$

$u^2 + 2u + 1 + v^2 = 4u^2 - 8u + 4 + 4v^2$

$0 = 3u^2 - 10u + 3 + 3v^2$

$3u^2 - 10u + 3v^2 + 3 = 0$

$3\!\left(u^2 - \frac{10}{3}u\right) + 3v^2 = -3$

$3\!\left(u - \frac{5}{3}\right)^2 - \frac{25}{3} + 3v^2 = -3$

$3\!\left(u - \frac{5}{3}\right)^2 + 3v^2 = \frac{16}{3}$

$\left(u - \frac{5}{3}\right)^2 + v^2 = \frac{16}{9}$

This is a circle with centre $\left(\dfrac{5}{3}, 0\right)$ and radius $\dfrac{4}{3}$ in the $w$-plane.

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Problem 7. Express $\dfrac◆LB◆(1+i)^6◆RB◆◆LB◆(1-i\sqrt{3})^4◆RB◆$ in the form $a + bi$.

Hint

Write each term in exponential form and use the laws of indices.

Answer

$1 + i = \sqrt{2}\,e^{i\pi/4}$ and $1 - i\sqrt{3} = 2\,e^{-i\pi/3}$.

$\frac◆LB◆(1+i)^6◆RB◆◆LB◆(1-i\sqrt{3})^4◆RB◆ = \frac◆LB◆(\sqrt{2}\,e^{i\pi/4})^6◆RB◆◆LB◆(2\,e^{-i\pi/3})^4◆RB◆ = \frac◆LB◆8e^{3\pi i/2}◆RB◆◆LB◆16\,e^{-4\pi i/3}◆RB◆ = \frac{1}{2}\,e^{i(3\pi/2 + 4\pi/3)}$

$3\pi/2 + 4\pi/3 = \frac◆LB◆9\pi + 8\pi◆RB◆◆LB◆6◆RB◆ = \frac◆LB◆17\pi◆RB◆◆LB◆6◆RB◆ = 2\pi + \frac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆$

So: $\dfrac{1}{2}\,e^{5\pi i/6} = \dfrac{1}{2}\!\left(\cos\dfrac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆ + i\sin\dfrac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆\right) = \dfrac{1}{2}\!\left(-\dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ + \dfrac{1}{2}\,i\right) = -\dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆4◆RB◆ + \dfrac{1}{4}\,i$

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Problem 8. The transformation $T$ from the $z$-plane to the $w$-plane is given by $w = z^2$. The region $R$ in the $z$-plane is defined by $1 \leq |z| \leq 2$ and $0 \leq \arg(z) \leq \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$. Find and describe the image of $R$ under $T$.

Hint

Under $w = z^2$, the modulus squares and the argument doubles.

Answer

If $z = re^{i\theta}$, then $w = r^2 e^{2i\theta}$.

• Modulus: $1 \leq r \leq 2 \implies 1 \leq r^2 \leq 4$, so $1 \leq |w| \leq 4$.
• Argument: $0 \leq \theta \leq \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ \implies 0 \leq 2\theta \leq \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$, so $0 \leq \arg(w) \leq \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$.

The image is the region in the first quadrant of the $w$-plane between the circles $|w| = 1$ and $|w| = 4$, bounded by the rays $\arg(w) = 0$ and $\arg(w) = \dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$.

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Problem 9. Solve the equation $z^3 + z^2 + z + 1 = 0$ by recognising it as a geometric series, and hence show that $\cos\dfrac◆LB◆2\pi◆RB◆◆LB◆4◆RB◆ + \cos\dfrac◆LB◆4\pi◆RB◆◆LB◆4◆RB◆ + \cos\dfrac◆LB◆6\pi◆RB◆◆LB◆4◆RB◆ = -1$.

Hint

Factor $z^3 + z^2 + z + 1 = (z+1)(z^2+1)$. The roots are the 4th roots of unity excluding $z = 1$.

Answer

$z^3 + z^2 + z + 1 = \dfrac{z^4 - 1}{z - 1} = 0 \implies z^4 = 1$ with $z \neq 1$.

The 4th roots of unity are $1, i, -1, -i$, so the solutions are $z = i, -1, -i$.

Equivalently, the roots are $e^{k\pi i/2}$ for $k = 1, 2, 3$.

The sum of roots (by Vieta, coefficient of $z^2$ divided by leading coefficient) is $-1$:

$i + (-1) + (-i) = -1 \quad \checkmark$

Now: $e^{i\pi/2} + e^{i\pi} + e^{3i\pi/2} = i + (-1) + (-i) = -1$.

Separating real and imaginary parts: $\cos\dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + \cos\pi + \cos\dfrac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ = -1$, i.e. $0 - 1 + 0 = -1$. ✓

Alternatively, the claim as stated uses $\cos\dfrac◆LB◆2\pi◆RB◆◆LB◆4◆RB◆ + \cos\dfrac◆LB◆4\pi◆RB◆◆LB◆4◆RB◆ + \cos\dfrac◆LB◆6\pi◆RB◆◆LB◆4◆RB◆ = \cos\dfrac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + \cos\pi + \cos\dfrac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ = 0 + (-1) + 0 = -1$. ✓ $\square$

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Problem 10. (a) Show that $\dfrac◆LB◆1◆RB◆◆LB◆e^{i\theta} - 1◆RB◆ = -\dfrac{1}{2} - \dfrac{i}{2}\cot\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆$ for $\theta \notin 2\pi\mathbb{Z}$.

(b) Hence, or otherwise, find $\displaystyle\sum_{k=1}^{n-1}\frac◆LB◆1◆RB◆◆LB◆1 - \omega^k◆RB◆$ where $\omega = e^{2\pi i/n}$.

Hint

For (a), multiply numerator and denominator by the conjugate $e^{-i\theta} - 1$ and use half-angle identities. For (b), use the result from (a) with $\theta = 2k\pi/n$.

Answer

(a) $\dfrac◆LB◆1◆RB◆◆LB◆e^{i\theta} - 1◆RB◆ = \dfrac◆LB◆e^{-i\theta} - 1◆RB◆◆LB◆(e^{i\theta} - 1)(e^{-i\theta} - 1)◆RB◆ = \dfrac◆LB◆e^{-i\theta} - 1◆RB◆◆LB◆2 - (e^{i\theta} + e^{-i\theta})◆RB◆ = \dfrac◆LB◆e^{-i\theta} - 1◆RB◆◆LB◆2 - 2\cos\theta◆RB◆$.

Numerator: $e^{-i\theta} - 1 = \cos\theta - 1 - i\sin\theta = -2\sin^2\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ - 2i\sin\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\cos\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = -2\sin\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\!\left(\sin\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ + i\cos\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\right)$.

Denominator: $2 - 2\cos\theta = 4\sin^2\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆$.

$\frac◆LB◆1◆RB◆◆LB◆e^{i\theta} - 1◆RB◆ = \frac◆LB◆-2\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\!\left(\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ + i\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\right)◆RB◆◆LB◆4\sin^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆ = \frac◆LB◆-\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ - i\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆2\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆ = -\frac{1}{2} - \frac{i}{2}\cot\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ \quad \square$

(b) Using (a): $\dfrac◆LB◆1◆RB◆◆LB◆1 - \omega^k◆RB◆ = -\dfrac◆LB◆1◆RB◆◆LB◆e^{2k\pi i/n} - 1◆RB◆ = \dfrac{1}{2} + \dfrac{i}{2}\cot\dfrac◆LB◆k\pi◆RB◆◆LB◆n◆RB◆$.

$\sum_{k=1}^{n-1}\frac◆LB◆1◆RB◆◆LB◆1 - \omega^k◆RB◆ = \sum_{k=1}^{n-1}\!\left(\frac{1}{2} + \frac{i}{2}\cot\frac◆LB◆k\pi◆RB◆◆LB◆n◆RB◆\right) = \frac{n-1}{2} + \frac{i}{2}\sum_{k=1}^{n-1}\cot\frac◆LB◆k\pi◆RB◆◆LB◆n◆RB◆$

The cotangent sum is zero by symmetry: $\cot\dfrac◆LB◆k\pi◆RB◆◆LB◆n◆RB◆ = -\cot\dfrac◆LB◆(n-k)\pi◆RB◆◆LB◆n◆RB◆$, so terms cancel in pairs.

Therefore: $\displaystyle\sum_{k=1}^{n-1}\frac◆LB◆1◆RB◆◆LB◆1 - \omega^k◆RB◆ = \frac{n - 1}{2}$.

:::

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8. Advanced Worked Examples

Example 8.1: De Moivre's theorem for $\cos 5\theta$

Problem. Express $\cos 5\theta$ in terms of $\cos\theta$.

Solution. By de Moivre: $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$.

Expanding by the binomial theorem:

$(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta$

Equating real parts:

$\boxed{\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta}$

Example 8.2: Solving $z^6 = -64$

Problem. Solve $z^6 = -64$, giving answers in exponential form.

Solution. $-64 = 64e^{i\pi}$. The 6th roots are:

$z_k = 64^{1/6} \exp\!\left(\frac◆LB◆i(\pi + 2k\pi)◆RB◆◆LB◆6◆RB◆\right) = 2\exp\!\left(\frac◆LB◆i(2k+1)\pi◆RB◆◆LB◆6◆RB◆\right)$

for $k = 0, 1, 2, 3, 4, 5$.

$z_0 = 2e^{i\pi/6}$, $z_1 = 2e^{i\pi/2}$, $z_2 = 2e^{5i\pi/6}$, $z_3 = 2e^{7i\pi/6}$, $z_4 = 2e^{3i\pi/2}$, $z_5 = 2e^{11i\pi/6}$.

These lie on a circle of radius 2, at angles $30°, 90°, 150°, 210°, 270°, 330°$.

Example 8.3: Loci — perpendicular bisector

Problem. Find the Cartesian equation of the locus $|z - 3 - 4i| = |z + 1 - 2i|$.

Solution. Let $z = x + iy$.

$|z - (3+4i)| = |z - (-1+2i)|$

$\sqrt{(x-3)^2 + (y-4)^2} = \sqrt{(x+1)^2 + (y-2)^2}$

Squaring: $(x-3)^2 + (y-4)^2 = (x+1)^2 + (y-2)^2$.

$x^2 - 6x + 9 + y^2 - 8y + 16 = x^2 + 2x + 1 + y^2 - 4y + 4$

$-8x - 4y + 20 = 0 \implies \boxed{2x + y = 5}$

This is the perpendicular bisector of the segment joining $3+4i$ and $-1+2i$.

Example 8.4: Region defined by an inequality

Problem. Shade on an Argand diagram the region defined by $|z - 2i| \leq 3$ and $0 \leq \arg(z) \leq \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$.

Solution. $|z - 2i| \leq 3$ is the closed disc of radius 3 centred at $2i$ (i.e., $(0, 2)$).

$0 \leq \arg(z) \leq \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$ is the sector between the positive real axis and the line $y = x$ (for $x \geq 0$).

The required region is the intersection: a segment of the disc in the first quadrant between angles $0$ and $\pi/4$.

Example 8.5: Complex transformation — rotation and enlargement

Problem. The transformation $T$ maps the complex plane by $w = (1+i)z + 2i$. Describe $T$ fully and find the image of the line $\mathrm{Re}(z) = 1$.

Solution. $w = (1+i)z + 2i = \sqrt{2}\,e^{i\pi/4}\,z + 2i$.

$T$ is an enlargement by scale factor $\sqrt{2}$, rotation by $45°$ anticlockwise about the origin, followed by a translation by $2i$.

For $\mathrm{Re}(z) = 1$: $z = 1 + it$. $w = (1+i)(1+it) + 2i = 1 + it + i - t + 2i = (1-t) + i(3+t)$.

$\mathrm{Re}(w) = 1-t$, $\mathrm{Im}(w) = 3+t$. Eliminating $t$: $\mathrm{Im}(w) = 3 + (1 - \mathrm{Re}(w)) = 4 - \mathrm{Re}(w)$.

The image is the line $\boxed{u + v = 4}$ (where $w = u + iv$).

Example 8.6: Sum of roots of unity

Problem. Show that $\displaystyle\sum_{k=0}^{n-1} \omega^k = 0$ where $\omega = e^{2\pi i/n}$.

Solution. This is a geometric series with ratio $\omega \neq 1$:

$\sum_{k=0}^{n-1} \omega^k = \frac◆LB◆1 - \omega^n◆RB◆◆LB◆1 - \omega◆RB◆ = \frac◆LB◆1 - 1◆RB◆◆LB◆1 - \omega◆RB◆ = 0$

$\blacksquare$

Example 8.7: Product of roots of unity

Problem. Show that the product of all $n$-th roots of unity is $(-1)^{n+1}$.

Solution. The $n$-th roots of unity are the roots of $z^n - 1 = 0$. By Vieta's formulae, the product of all roots is $(-1)^n \times (-1) = (-1)^{n+1}$.

Alternatively: the roots are $1, \omega, \omega^2, \ldots, \omega^{n-1}$, so the product is $\omega^{0+1+2+\cdots+(n-1)} = \omega^{n(n-1)/2} = e^{\pi i(n-1)} = (-1)^{n-1} = (-1)^{n+1}$. $\blacksquare$

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9. Common Pitfalls

| Pitfall | Correct Approach | | ----------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------ | -------- | --- | --- | --- | --- | --- | --- | --------------------------------------------------- | --- | ---- | --- | --- | --- | --- | | Forgetting that $\arg(z)$ is measured from the positive real axis | $\arg(z)$ is the angle anticlockwise from the positive $x$-axis, range $(-\pi, \pi]$ or $[0, 2\pi)$ | | Confusing $| z-w |$ with $| z | - | w |$ | $| z-w |$ is the distance between $z$ and $w$; in general $| z-w | \neq | z | - | w |$ | | Missing roots when solving $z^n = w$ | There are always exactly $n$ distinct roots; check your $k$ values cover $0$ to $n-1$ | | Incorrectly applying de Moivre to non-integer powers | De Moivre's theorem $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ holds for integer $n$ only |

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10. Additional Exam-Style Questions

Question 8

Solve $z^4 = 8(1 + i\sqrt{3})$, giving roots in the form $r(\cos\theta + i\sin\theta)$.

Solution

$8(1+i\sqrt{3}) = 16e^{i\pi/3}$.

$z_k = 2\exp\!\left(\dfrac◆LB◆i(\pi/3 + 2k\pi)◆RB◆◆LB◆4◆RB◆\right)$ for $k = 0, 1, 2, 3$.

$z_0 = 2(\cos 15° + i\sin 15°)$, $z_1 = 2(\cos 105° + i\sin 105°)$, $z_2 = 2(\cos 195° + i\sin 195°)$, $z_3 = 2(\cos 285° + i\sin 285°)$.

Question 9

Prove that $\cos^4\theta = \dfrac{3}{8} + \dfrac{1}{2}\cos 2\theta + \dfrac{1}{8}\cos 4\theta$.

Solution

$\cos^2\theta = \dfrac◆LB◆1+\cos 2\theta◆RB◆◆LB◆2◆RB◆$.

$\cos^4\theta = \left(\dfrac◆LB◆1+\cos 2\theta◆RB◆◆LB◆2◆RB◆\right)^{\!2} = \dfrac◆LB◆1 + 2\cos 2\theta + \cos^2 2\theta◆RB◆◆LB◆4◆RB◆$.

$\cos^2 2\theta = \dfrac◆LB◆1+\cos 4\theta◆RB◆◆LB◆2◆RB◆$.

$\cos^4\theta = \dfrac{1}{4} + \dfrac◆LB◆\cos 2\theta◆RB◆◆LB◆2◆RB◆ + \dfrac◆LB◆1+\cos 4\theta◆RB◆◆LB◆8◆RB◆ = \dfrac{3}{8} + \dfrac◆LB◆\cos 2\theta◆RB◆◆LB◆2◆RB◆ + \dfrac◆LB◆\cos 4\theta◆RB◆◆LB◆8◆RB◆$. $\blacksquare$

Question 10

The complex number $z$ satisfies $|z-1| = |z+1|$ and $|z-3i| = 3$. Find $z$.

Solution

$|z-1| = |z+1|$: perpendicular bisector of $1$ and $-1$, giving $\mathrm{Re}(z) = 0$. So $z = iy$.

$|z-3i| = 3 \implies |iy - 3i| = 3 \implies |y-3| = 3 \implies y - 3 = \pm 3$.

$y = 6$ or $y = 0$. So $z = 6i$ or $z = 0$.

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11. Connections to Other Topics

11.1 Complex numbers and matrices

Complex eigenvalues of 2×2 matrices correspond to rotation-scaling transformations. See Matrices.

11.2 Complex numbers and hyperbolic functions

$e^{ix} = \cos x + i\sin x$ connects exponential, trigonometric, and hyperbolic functions. See Hyperbolic Functions.

11.3 Complex numbers and polar coordinates

Argand diagrams and polar form $(r, \theta)$ connect to polar coordinates. See Polar Coordinates.

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12. Key Results Summary

| Result | Formula | | ----------------------------- | --------------------------------------------------------- | --- | -------------------------------- | --- | --- | | Modulus | $| z | = \sqrt{a^2+b^2}$ for $z = a+bi$ | | Argument | $\arg(z) = \arctan(b/a)$ (adjusting for quadrant) | | Euler's formula | $e^{i\theta} = \cos\theta+i\sin\theta$ | | De Moivre | $(\cos\theta+i\sin\theta)^n = \cos n\theta+i\sin n\theta$ | | $n$-th roots of unity | $z_k = e^{2\pi ik/n}$, $k = 0, \ldots, n-1$ | | Locus: circle | $| z-a | =r$ | | Locus: perpendicular bisector | $| z-a | = | z-b |$ | | Locus: half-line | $\arg(z-a) = \theta$ |

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13. Further Exam-Style Questions

Question 11

Solve $z^3 = -8i$, giving roots in Cartesian form.

Solution

$-8i = 8e^{-i\pi/2}$. Roots: $z_k = 2\exp\!\left(\dfrac◆LB◆-i\pi/2 + 2k\pi i◆RB◆◆LB◆3◆RB◆\right)$ for $k=0,1,2$.

$z_0 = 2e^{-i\pi/6} = 2\!\left(\dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ - \dfrac{i}{2}\right) = \sqrt{3}-i$.

$z_1 = 2e^{i\pi/2} = 2i$.

$z_2 = 2e^{7i\pi/6} = 2\!\left(-\dfrac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ - \dfrac{i}{2}\right) = -\sqrt{3}-i$.

$\boxed{z = \sqrt{3}-i,\; 2i,\; -\sqrt{3}-i}$

Question 12

Prove that $|z_1 z_2| = |z_1||z_2|$ for any complex numbers $z_1, z_2$.

Solution

Let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$.

$z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}$.

$|z_1 z_2| = r_1 r_2 = |z_1||z_2|$. $\blacksquare$

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14. Advanced Topics

14.1 The exponential form of complex numbers

Any non-zero complex number can be written as $z = re^{i\theta}$ where $r = |z|$ and $\theta = \arg(z)$.

This form makes multiplication and division particularly simple:

• $z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}$
• $z_1/z_2 = (r_1/r_2) e^{i(\theta_1-\theta_2)}$

14.2 Euler's identity

Setting $\theta = \pi$ in Euler's formula: $e^{i\pi} + 1 = 0$.

This connects five fundamental constants: $e$, $i$, $\pi$, $1$, and $0$.

14.3 Complex conjugate and roots of polynomials

If $P(z)$ is a polynomial with real coefficients and $z = a + bi$ is a root, then $\bar{z} = a - bi$ is also a root. This is because $\overline{P(z)} = P(\bar{z})$ for real-coefficient polynomials.

14.4 Regions in the Argand diagram

Inequality	Region
$\|z-a\| < r$	Interior of circle (open disc)
$\|z-a\| \leq r$	Closed disc
$\alpha < \arg(z-a) < \beta$	Sector (angular region)
$\mathrm{Re}(z) > k$	Half-plane to the right of $x = k$
$\mathrm{Im}(z) > k$	Half-plane above $y = k$

14.5 The exponential form of $\sin$ and $\cos$

From $e^{i\theta} = \cos\theta + i\sin\theta$ and $e^{-i\theta} = \cos\theta - i\sin\theta$:

$\cos\theta = \frac◆LB◆e^{i\theta}+e^{-i\theta}◆RB◆◆LB◆2◆RB◆, \qquad \sin\theta = \frac◆LB◆e^{i\theta}-e^{-i\theta}◆RB◆◆LB◆2i◆RB◆$

These are essential for deriving trigonometric identities and solving certain integrals.

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15. Further Exam-Style Questions

Question 13

Express $\dfrac◆LB◆1+e^{i\theta}◆RB◆◆LB◆1-e^{i\theta}◆RB◆$ in the form $a+bi$.

Solution

$\dfrac◆LB◆1+e^{i\theta}◆RB◆◆LB◆1-e^{i\theta}◆RB◆ = \dfrac◆LB◆e^{i\theta/2}(e^{-i\theta/2}+e^{i\theta/2})◆RB◆◆LB◆e^{i\theta/2}(e^{-i\theta/2}-e^{i\theta/2})◆RB◆ = \dfrac◆LB◆2\cos(\theta/2)◆RB◆◆LB◆-2i\sin(\theta/2)◆RB◆ = \dfrac◆LB◆i\cos(\theta/2)◆RB◆◆LB◆\sin(\theta/2)◆RB◆ = \boxed{i\cot(\theta/2)}$

Question 14

Prove that the sum of the $n$-th roots of unity is zero.

Solution

The $n$-th roots of unity are $1, \omega, \omega^2, \ldots, \omega^{n-1}$ where $\omega = e^{2\pi i/n}$.

This is a geometric series: $\displaystyle\sum_{k=0}^{n-1} \omega^k = \frac◆LB◆1-\omega^n◆RB◆◆LB◆1-\omega◆RB◆ = \frac◆LB◆1-1◆RB◆◆LB◆1-\omega◆RB◆ = 0$. $\blacksquare$

Question 15

Find all complex numbers $z$ such that $z\bar{z} + z + \bar{z} = 3$.

Solution

Let $z = x+iy$. Then $\bar{z} = x-iy$ and $z\bar{z} = x^2+y^2$.

$x^2+y^2 + 2x = 3 \implies (x+1)^2 + y^2 = 4$.

This is a circle with centre $(-1, 0)$ and radius $2$. All complex numbers on this circle satisfy the equation.

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16. Further Advanced Topics

16.1 The fundamental theorem of algebra

Every polynomial of degree $n \geq 1$ with complex coefficients has exactly $n$ roots (counting multiplicity) in the complex numbers.

This means: every polynomial can be factored as $P(z) = a(z - z_1)(z - z_2)\cdots(z - z_n)$.

16.2 Complex numbers as rotations and dilations

Multiplication by $re^{i\theta}$ represents:

• Dilation by scale factor $r$
• Rotation by angle $\theta$ anticlockwise

This provides a geometric interpretation of all complex arithmetic.

16.3 $n$-th roots of any complex number

To solve $z^n = w = re^{i\phi}$:

$z_k = r^{1/n} \exp\!\left(\frac◆LB◆i(\phi + 2k\pi)◆RB◆◆LB◆n◆RB◆\right) \quad \text{for } k = 0, 1, \ldots, n-1$

The roots lie on a circle of radius $r^{1/n}$, equally spaced.

16.4 The complex exponential function

$e^z = e^{x+iy} = e^x(\cos y + i\sin y)$ for $z = x + iy$.

Key properties:

• $|e^z| = e^x$
• $\arg(e^z) = y$ (mod $2\pi$)
• $e^{z_1+z_2} = e^{z_1}e^{z_2}$
• $e^z \neq 0$ for all $z$
• $e^z = 1 \iff z = 2k\pi i$ for some integer $k$

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17. Further Exam-Style Questions

Question 16

Find all complex numbers $z$ such that $z^4 = -16$.

Solution

$-16 = 16e^{i\pi}$. Roots: $z_k = 2\exp\!\left(\dfrac◆LB◆i(\pi+2k\pi)◆RB◆◆LB◆4◆RB◆\right)$ for $k=0,1,2,3$.

$z_0 = 2e^{i\pi/4} = \sqrt{2}+i\sqrt{2}$, $z_1 = 2e^{i3\pi/4} = -\sqrt{2}+i\sqrt{2}$,

$z_2 = 2e^{i5\pi/4} = -\sqrt{2}-i\sqrt{2}$, $z_3 = 2e^{i7\pi/4} = \sqrt{2}-i\sqrt{2}$.

Question 17

Prove that for any complex number $z$: $\overline{z_1 z_2} = \bar{z}_1 \cdot \bar{z}_2$.

Solution

Let $z_1 = a+bi$ and $z_2 = c+di$.

$z_1 z_2 = (ac-bd) + (ad+bc)i$.

$\overline{z_1 z_2} = (ac-bd) - (ad+bc)i$.

$\bar{z}_1 \cdot \bar{z}_2 = (a-bi)(c-di) = ac - adi - bci + bdi^2 = (ac-bd) - (ad+bc)i$.

Equal. $\blacksquare$

Question 18

The complex numbers $z$ and $w$ satisfy $|z| = 3$, $|w| = 4$, and $|z+w| = 5$. Find $|z-w|$.

Solution

$|z+w|^2 = |z|^2 + |w|^2 + 2\mathrm{Re}(z\bar{w}) = 9+16+2\mathrm{Re}(z\bar{w}) = 25$.

$\mathrm{Re}(z\bar{w}) = 0$.

$|z-w|^2 = |z|^2 + |w|^2 - 2\mathrm{Re}(z\bar{w}) = 9+16-0 = 25$.

$\boxed{|z-w| = 5}$

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18. Further Advanced Topics

18.1 Riemann surfaces and multi-valued functions

The complex logarithm, $n$-th root, and inverse trigonometric functions are all multi-valued. Branch cuts are used to define single-valued branches (principal values).

18.2 The complex plane and stereographic projection

The extended complex plane $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$ is topologically a sphere (the Riemann sphere). Stereographic projection maps each point on the sphere (except the north pole) to a unique point in the complex plane.

18.3 Complex analysis connections

While complex analysis (Cauchy's theorem, residue calculus) is beyond A-Level, the fundamental concepts appear:

• Cauchy's integral formula: $f(a) = \dfrac◆LB◆1◆RB◆◆LB◆2\pi i◆RB◆\displaystyle\oint_C \frac{f(z)}{z-a}\,dz$
• Residue theorem: $\displaystyle\oint_C f(z)\,dz = 2\pi i \sum \text{Res}(f, a_k)$

These are mentioned for context and further study.

18.4 De Moivre's theorem — number theory applications

De Moivre's theorem connects complex numbers to number theory:

• Fermat's theorem on sums of two squares: $p \equiv 1 \pmod 4 \implies p = a^2 + b^2$
• Wilson's theorem: $(p-1)! \equiv -1 \pmod p$ for prime $p$

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19. Further Exam-Style Questions

Question 19

Express $\cos 5\theta + i\sin 5\theta$ in terms of $\cos\theta$ and $\sin\theta$ using the binomial theorem.

Solution

$(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta$.

Real part: $\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$.

Using $\sin^2\theta = 1-\cos^2\theta$:

$= \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-\cos^2\theta)^2$

$= \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta$

$= \boxed{16\cos^5\theta - 20\cos^3\theta + 5\cos\theta}$.

Question 20

Prove that $|z+w|^2 + |z-w|^2 = 2(|z|^2+|w|^2)$ for all complex numbers $z, w$ (the parallelogram law).

Solution

$|z+w|^2 + |z-w|^2 = (z+w)(\bar{z}+\bar{w}) + (z-w)(\bar{z}-\bar{w})$

$= |z|^2+z\bar{w}+w\bar{z}+|w|^2 + |z|^2-z\bar{w}-w\bar{z}+|w|^2 = 2|z|^2 + 2|w|^2$.

This is the parallelogram law: the sum of squares of the diagonals equals the sum of squares of all four sides. $\blacksquare$