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.

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.

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 $(a, b)$.

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 $\theta = \arg(z)$.

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

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

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

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

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

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:

Using the compound angle identities. ✓

For negative integers, note that $\dfrac{1}{\cos\theta + i\sin\theta} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$ 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 multiplies its angle by $n$. This is because multiplication of complex numbers Adds their arguments: $r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)}$.

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:

Equating real and imaginary parts:

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

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

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

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

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:

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

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

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:

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

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

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

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.

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 $e^x$:

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

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

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

Setting $\theta = \pi$:

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

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

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 $\theta$.

The standard laws of indices extend :

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

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.

$|z - a| = r \iff \sqrt{(x - \alpha)^2 + (y - \beta)^2} = r \iff (x - \alpha)^2 + (y - \beta)^2 = r^2$

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.

This locus represents all points equidistant from $a$ and $b$Which is the perpendicular bisector Of the line segment joining $a$ and $b$.

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 $a$.

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

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

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 $w$).

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 $b$.

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.

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).

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

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

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

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

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

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

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

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

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

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

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 $\cos\dfrac{2\pi}{5} + \cos\dfrac{4\pi}{5} = -\dfrac{1}{2}$.

Problem 5. Sketch on separate Argand diagrams the loci given by (a) $|z - 1 - i| = |z - 3 + i|$ And (b) $\arg(z - 2) = \dfrac{\pi}{3}$. Find the complex number(s) satisfying both Conditions simultaneously.

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

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

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 $0 \leq \arg(z) \leq \dfrac{\pi}{4}$. Find and describe the image of $R$ under $T$.

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