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.

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for almost two decades. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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Deriving the polar form for complex number

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