Lecture 3

How can we take the square root of the imaginary unit? Consider the equivalent question of finding the complex number \(z\) who, when multiplied by itself, returns \(i\):\[z^2=i\] We will write \(z\) in terms of its real and imaginary components: \[z=x+iy \implies (x+iy)^2=i\] and expand: \[x^2-y^2+i~2xy=i\] Notice we now have an expression for \(i\) of form \(a+ib\), so we can see \[\begin{align}\Re[i]&=0=x^2-y^2\\\Im[i]&=1=2xy\end{align}\] which allows us to solve for \(x\) and \(y\): \[\begin{align}x&=\pm y\\ 2y^2&=1\\\implies y&=\pm\frac{\sqrt{2}}{2}\\x&=y\end{align}\]

And so we arrive at our solution⊕ The two square roots of \(i\) visualized on the unit circle.: \[\begin{align}\sqrt{i}=z&=x+iy\\&=\boxed{\pm\frac{\sqrt{2}}{2}\left(1+i\right)}\end{align}\]

Vectors

Graphically, complex addition works the same as vector addition:

Polar Coordinates

Working with complex numbers in Cartesian is tedious. We will discover that polar coordinates are a much more natural choice. How can we describe \(z\) in terms of \(r\) and \(\theta\)? In other words, how can we find:\[z(r(x,y),\theta(x,y)) =~?\] Shown graphically:

Trigonometry makes this quite easy: \[\begin{align}r&=\sqrt{x^2+y^2}\\\theta&=\tan^{-1}\left(\frac{y}{x}\right)\end{align}\] Notice our expression for \(r\) is equivalent to \(|z|\), the modulusAlso called the magnitude of \(z\). One may then expect that our expression for \(\theta\) ought also be named, and indeed it is — the function \(\arg(z)=\theta\) is called the argumentAlso called the phase of \(z\).

Now we can express \(x\) and \(y\) in terms of \(r\) and \(\theta\):\[\begin{align}x&=r\cos\theta\\ y&=r\sin\theta\\\implies z&=x+iy=r\cos\theta+i~r\sin\theta\\ &=r\left[\cos\theta+i\sin\theta\right]\\ &=re^{i\theta}\end{align}\] Beauty.

Theorem: Euler's formula \[e^{i\theta}=\cos\theta+i\sin\theta\]⊕ When taken as a function, the expression is sometimes denoted \(\mathrm{cis}~\theta\), from "cosine plus i sine"

Definition. The sets \(\mathbb{E}\) and \(\mathbb{O}\) denote the even and odd naturals, respectively: \[ \begin{align} \mathbb{E} &= \{n\in\mathbb{N}~|~(\exists k)[k\in\mathbb{N}\land n = 2k]\}\\ \mathbb{O} &= \mathbb{N}\backslash\mathbb{E} \end{align} \]

Proof. Using power series:⊕ \[\boxed{\begin{align}\mathbf{Recall.}\\\\ e^x &= \sum_{n=0}^{\infty}\frac{x^n}{n!} \\\\ \cos x &= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n!} \\\\ \sin x &= \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}~~~~~\\& \end{align}}\]

\[ \begin{align} e^{i\theta}&=\sum_{n=0}^{\infty}\frac{(i\theta)^n}{n!} = \sum_{n=0}^{\infty}\frac{i^n\theta^n}{n!}\\ &= \sum_{n\in\mathbb{E}}\frac{i^n\theta^n}{n!}+\sum_{n\in\mathbb{O}}\frac{i^n\theta^n}{n!}\\ &= \sum_{m=0}^\infty\frac{i^{2m}\theta^{2m}}{2m!}+\sum_{m=0}^\infty\frac{i^{2m+1}\theta^{2m+1}}{(2m+1)!}\\ &= \sum_{m=0}^\infty\frac{(-1)^m\theta^{2m}}{2m!} + \sum_{m=0}^\infty\frac{i(-1)^m\theta^{2m+1}}{(2m+1)!}\\ &= \sum_{m=0}^\infty\frac{(-1)^m\theta^{2m}}{2m!} + i\sum_{m=0}^\infty\frac{(-1)^m\theta^{2m+1}}{(2m+1)!}\\ &= \cos\theta + i\sin\theta~~\square \end{align} \]

Complex multiplication in polar form

Let us take two complex numbers \(z_1\) and \(z_2\) of the form \(z_k = r_ke^{i\theta_k}\). Their product takes the form \[z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}\]

Two important observations can be made from this result. First, when multiplying complex numbers, the modulus of the product is equal to the product of the moduli of the multiplicands: \[\left|\prod_kz_k\right|=\prod_k|z_k|\] Second, the argument of the product is equal to the sum of the arguments of the multiplicands: \[\arg \prod_kz_k = \sum_k\arg z_k\]

These properties make the multiplication of many complex numbers incredibly simple.