Theory

Theory 1

Multiplication of complex numbers is much easier to understand when the numbers are written using polar form.

There is a shorthand ‘$\mathrm{cis}$’ notation. Convert to polar coordinates, so $a=r\cos \theta$ and $b=r\sin \theta$:

$$\begin{array}{cc}a+bi& \gg \gg r\cos \theta +r\sin \theta i\\ & \\ & \gg \gg r(\cos \theta +i\sin \theta )\\ & \\ & \gg \gg r\mathrm{cis}\theta \end{array}$$

The “$\mathrm{cis}$” stands for $\cos \theta +i\sin \theta$. For example:

$$\begin{array}{c}\sqrt{2}-\sqrt{2}i\gg \gg 2(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\\ \\ \gg \gg 2\cos (-\frac{\pi }{4})+2\sin (-\frac{\pi }{4})i\\ \\ \gg \gg 2\mathrm{cis}(-\frac{\pi }{4})\end{array}$$

---

Euler Formula

General Euler Formula:

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

On the unit circle $r=1$:

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

The form $r{e}^{i\theta }$ expresses the same data as the $\mathrm{cis}$ form. The principal advantage of the form $r{e}^{i\theta }$ is that it reveals the rule for multiplication, which comes from exponent laws:

Complex multiplication - Exponential form

$$r_1{e}^{i{\theta }_1}\cdot r_2{e}^{i{\theta }_2}=(r_1{r}_2)e^{i({\theta }_1+{\theta }_2)}$$

In words:

• Multiply radii
• Add angles

Notice:

$$\text{multiply by }{e}^{i\frac{\pi }{2}}⟺\text{rotate by }+{90}^{\circ }$$

Notice:

$$e^{i\frac{\pi }{2}}=+i$$

Therefore $i$ ‘acts upon’ other numbers by rotating them ${90}^{\circ }$ counterclockwise!

---

De Moivre’s Theorem - Complex powers

In exponential notation:

$$(r{e}^{i\theta }{)}^n=r^n{e}^{i\cdot n\theta }$$

In $\mathrm{cis}$ notation:

$$(r\mathrm{cis}\theta {)}^n=r^n\mathrm{cis}(n\theta )$$

Expanded $\mathrm{cis}$ notation:

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

So the power of $n$ acts like this:

• Stretch: $r$ to $r^n$
• Rotate: by $n$ increments of $\theta$

---

Extra - Derivation of Euler Formula

Recall the power series for $e^x$:

$$e^x=1+\frac{1}{1!}x+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+\dots =\sum _{i=0}^{\infty }\frac{1}{i!}{x}^i$$

Plug in $x=i\theta$:

$$\begin{array}{c}{e}^{i\theta }\gg \gg 1+(i\theta )+\frac{1}{2!}{(i\theta )}^2+\frac{1}{3!}{(i\theta )}^3+\cdots +\end{array}$$

Simplify terms:

$$\gg \gg 1+i\theta -\frac{1}{2!}{\theta }^2-\frac{1}{3!}i{\theta }^3+\frac{1}{4!}{\theta }^4+\frac{1}{5!}i{\theta }^5-\frac{1}{6!}{\theta }^6-\frac{1}{7!}i{\theta }^7+\frac{1}{8!}{\theta }^8+\cdots$$

Separate by $i$-factor. Select out the $\overbrace{\text{terms with }i}$:

$$\gg \gg 1\overbrace{+i\theta }-\frac{1}{2!}{\theta }^2\overbrace{-\frac{1}{3!}i{\theta }^3}+\frac{1}{4!}{\theta }^4\overbrace{+\frac{1}{5!}i{\theta }^5}-\frac{1}{6!}{\theta }^6\overbrace{-\frac{1}{7!}i{\theta }^7}+\frac{1}{8!}{\theta }^8+\cdots$$

Separate into a series without $i$ and a series with $i$:

$$\gg \gg (1-\frac{1}{2!}{\theta }^2+\frac{1}{4!}{\theta }^4-\cdots )+(\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\cdots )i$$

Identify $\cos \theta$ and $\sin \theta i$. Write trig series:

$$\begin{array}{cc}\cos \theta & =1-\frac{1}{2!}{\theta }^2+\frac{1}{4!}{\theta }^4-\cdots \\ & \\ \sin \theta & =\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\cdots \end{array}$$

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