Let we have $f(z)=u(r,\theta)+iv(r,\theta)$ where $z=re^{i\theta}$ (being $r\neq{0}$). We have to prove that the Cauchy-Riemann equations in polar coordinates are defined in the following way: $\space$
$ru_{r}=v_{\theta}$ ; $rv_{r}=-u_{\theta}$
I have started doing the exercise but I have no idea how to continue with the prove... This is the only thing I have done: $\space$
$f(z)=f(re^{i\theta})=u(r,\theta)+iv(r,\theta)$ $\space$
Could someone help me with it?
The accompanying theorem says that a complex function is complex differentiable iff it is real differentiable and satisfies the Cauchy-Riemann equations. The real differentiability allows us to play around with the chain rule. Consider the function $\varphi:(0,\infty)\times(0,2\pi)\to\mathbb C,~(r,\theta)\mapsto r\mathrm e^{\mathrm i\theta}=r(\cos\theta+\mathrm i\sin\theta$. Then $f(r\mathrm e^{\mathrm i\theta})=f\circ\varphi$. We can calculate its Jacobian according to the chain rule:
$$\begin{align} \mathrm D(f\circ\varphi)&=\mathrm Df(\varphi)\mathrm D\varphi\\ &=\begin{pmatrix}u_x(r\mathrm e^{\mathrm i\theta})&u_y(r\mathrm e^{\mathrm i\theta})\\v_x(r\mathrm e^{\mathrm i\theta})&v_y(\mathrm e^{\mathrm i\theta})\end{pmatrix}\begin{pmatrix}\cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta\end{pmatrix}. \end{align}$$
You can carry out the multiplication, apply the Cauchy-Riemann equations and simplify the matrix, then consider that the Jacobian $\mathrm D(f\circ\varphi)$ contains exactly the partial derivatives wrt $r,\theta$. You can now read off the polar version of the C-R equations from the matrix.
