MTH2003 Complex Analysis Pre-Lecture 1

• Recording.
• Lecture Notes.
• Lecture: MTH2003 Complex Analysis Lecture 1.

Complex numbers are defined as: $z=x+iy$, where $x,y\in \mathbb{R}$, $i^2=-1$. These are often separated into real/Imaginary Parts $\text{Re}(z)=x$, $\text{Im}(z)=y$.

This set $\mathbb{C}$ is not just a two-dimensional vector space (with vector addition and scalar multiplication), but a field (with multiplication and division).

• Addition: $z_1+z_2=(x_1+x_2)+i(y_1+y_2)$.
• Multiplication: $z_1{z}_2=(x_1{x}_2-y_1{y}_2)+i(x_1{y}_2+x_2{y}_1)$.
• Division: Multiply by conjugate $\overline{z}=x-iy$; result scaled by $|z{|}^2$ (modulus: $|z|=\sqrt{x^2+y^2}$).

It also has two useful properties:

• Triangle Inequality: $|z_1{z}_2|=|z_1||z_2|$, $|z_1+z_2|\le |z_1|+|z_2|$.
• Part Separation: $z+\overline{z}=2\text{Re}(z)$, $z-\overline{z}=2i\text{Im}(z)$.

Since complex numbers are a 2D field, we can plot them by separating their real and imaginary parts. This is called an Argand diagram, and it is normally written in polar coordinates:

• Polar Form: $z=r(\cos \theta +i\sin \theta )$, where $r=|z|$, $\theta =\arg (z)$ — the modulus and argument of $z$.
• Exponential Form: $z=r{e}^{i\theta }$, where $e^{i\theta }=\cos \theta +i\sin \theta$ (Euler’s Formula).
• Euler’s Identity: $e^{i\pi }+1=0$.

This exponential form is very useful for solving complex algebraic equations, for example:

$$\begin{array}{rl}\text{Solve }{Z}^3& =8i\\ \text{Express }8i\text{ in polar form: }& 8i=8\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)=8e^{i\frac{\pi }{2}}.\\ \text{General roots: }Z& =\sqrt[3]{8}\cdot e^{i\left(\frac{\pi /2+2\pi k}{3}\right)},k=0,1,2.\\ \text{Simplify: }Z& =2e^{i\left(\frac{\pi }{6}+\frac{2\pi k}{3}\right)}.\\ \text{Roots: }k=0:& 2\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)=\sqrt{3}+i,\\ k=1:& 2\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)=-\sqrt{3}+i,\\ k=2:& 2\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)=-2i.\end{array}$$

We can simplify this process of finding the roots of complex numbers using De Moivre’s Theorem: $z^n=r^n(\cos n\theta +i\sin n\theta )$. With this, we can then calculate the $n$-th roots of unity, fairly simply:

$$\begin{array}{rl}\text{Solve }{Z}^n& =1.\\ \text{Write }1\text{ in polar form: }& 1=e^{i2\pi k}(k\in \mathbb{Z}).\\ \text{Roots: }Z& =e^{i\frac{2\pi k}{n}},k=0,1,\dots ,n-1.\\ \text{Example (n=4): }& Z=e^{i0},e^{i\frac{\pi }{2}},e^{i\pi },e^{i\frac{3\pi }{2}}=1,i,-1,-i.\end{array}$$

These roots of unity form a cyclic group under multiplication, generated by a primitive root $\omega =e^{i\frac{2\pi }{n}}$. The group structure is $\{1,\omega ,{\omega }^2,\dots ,{\omega }^{n-1}\}$, where multiplying any two elements corresponds to rotating around the unit circle in the complex plane.

To prove De Moivre’s Theorem for $z=r(\cos \theta +i\sin \theta )$ and integer $n$, we use induction and Euler’s formula:

$$\begin{array}{rl}\text{Base case (n=1):}& z^1=r(\cos \theta +i\sin \theta ).\text{True.}\\ \text{Assume true for n=k:}& z^k=r^k(\cos k\theta +i\sin k\theta ).\\ \text{Prove for n=k+1:}& z^{k+1}=z\cdot z^k=r(\cos \theta +i\sin \theta )\cdot r^k(\cos k\theta +i\sin k\theta )\\ & =r^{k+1}\left[\cos \theta \cos k\theta -\sin \theta \sin k\theta +i(\sin \theta \cos k\theta +\cos \theta \sin k\theta )\right]\\ & =r^{k+1}\left[\cos (k+1)\theta +i\sin (k+1)\theta \right].\text{(Trigonometric identities)}\\ \text{Conclusion:}& \text{By induction, }{z}^n=r^n(\cos n\theta +i\sin n\theta )\text{ for all }n\in \mathbb{N}.\\ \text{Extension via Euler:}& z=r{e}^{i\theta }⟹z^n=r^n{e}^{in\theta }=r^n(\cos n\theta +i\sin n\theta ).\square \end{array}$$