Differentiating polar functions using complex numbers

Question

I was wondering, given some polar function $r(\theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?

For example take the polar function $r=\cos(a\theta)$, also known as a rose curve for $a\in\mathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?

Answer 1

Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:

$ \cos{(ax)}=\frac{e^{iax}+e^{-iax}}{2} $,

$ \sin{(ax)}=\frac{e^{iax}-e^{-iax}}{2i} $.

Answer 2

Complex numbers are represented in the form $$ z= a e^{i\theta} $$ So real part of $ e^{iax} = \cos (ax) $ from Euler's identity is applicable here.

Answer 3

Let $f:\mathbb{R}\to\mathbb{R}$. Furthermore, let $r(f) = \cos(f) \:\mathrm{or} \:\sin(f)$. Form Eulers formula we know $$e^{if} = \cos f+i\sin f.$$ We know that $\operatorname{Re}e^{if} = \cos f$ and $\operatorname{Im}e^{if} = \sin f$.

Since derivative and real/imaginary part commute it is allowed to do what you have stated.