# Differentiating polar functions using complex numbers

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?

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.
$$\cos{(ax)}=\frac{e^{iax}+e^{-iax}}{2}$$,
$$\sin{(ax)}=\frac{e^{iax}-e^{-iax}}{2i}$$.
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 $$\operatorname{Im}e^{if} = \sin f$$.