Deriving the polar form for complex number

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.

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