I always stress to students the importance of the basics of complex numbers, that is

$z$
$=x+iy$
$=re^{i \theta}$
$=r(cos{\theta}+isin{\theta})$

The three forms of the complex numbers. I have shown how to derive from the cartesian ($x+iy$) to trigonometric form ($r(cos{\theta}+isin{\theta})$) form. But I seldom show students how to derive the polar form due to the rigor. So here we go. Lets first identify a few formulas from MF15.

$e^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...$

$cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...$

$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$

Let us consider the polar form then.

$e^{i\theta}=1+{i\theta}+\frac{{i\theta}^2}{2!}+\frac{{i\theta}^3}{3!}+\frac{{i\theta}^4}{4!}+\frac{{i\theta}^5}{5!}+...$

Resolving the $i's$

$e^{i\theta}=1+{i\theta}-\frac{{\theta}^2}{2!}-\frac{i{\theta}^3}{3!}+\frac{{\theta}^4}{4!}+\frac{i{\theta}^5}{5!}+...$

Rearranging them

$e^{i\theta}=1-\frac{{\theta}^2}{2!}+\frac{{\theta}^4}{4!}+...+{i\theta}-\frac{i{\theta}^3}{3!}+\frac{i{\theta}^5}{5!}+...$

Notice if we factorise $i$ out

$e^{i\theta}=1-\frac{{\theta}^2}{2!}+\frac{{\theta}^4}{4!}+...+i[{\theta}-\frac{{\theta}^3}{3!}+\frac{{\theta}^5}{5!}]+...$

Then with the formula we started with,

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

Voila! We are done and we have successfully derive our trigonometric form.