Complex Numbers Exponential Form

Given a complex number in cartesian form $z=r\cos \theta +ir\sin \theta$ and Euler’s formula:

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

we can express a complex number in its exponential form

$z=i{e}^{i\theta }$

Proof of Euler’s Formula

→ can be proved using taylor expansion

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\dots$

When $x$ is replaced by $ix$, the taylor series for $e^{ix}$ becomes:

$e^{ix}=1+ix+\frac{(ix{)}^2}{2!}+\frac{(ix{)}^3}{3!}+\frac{(ix{)}^4}{4!}\dots$

Simplify using properties of $i$:

$e^{ix}=1+ix-\frac{x^2}{2!}-\frac{i{x}^3}{3!}+\frac{x^4}{4!}+\frac{i{x}^5}{5!}+\dots$

The series becomes:

$e^{ix}=\underset{\cos x}{\underbrace{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \right)}}+i\underset{\sin x}{\underbrace{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\dots \right)}}$

$\therefore e^{ix}=\cos x+i\sin x■$