What you want here is the polar-form of the Cauchy-Riemann equations.
Now \(\displaystyle z = re^{i\theta } \quad \Rightarrow \quad x = r\cos (\theta )\quad \& \quad y = r\sin (\theta ).\)
From calculus we get:
\(\displaystyle \L \frac{{\partial u}}{{\partial r}} = \frac{{\partial u}}{{\partial x}}\frac{{\partial x}}{{\partial r}} + \frac{{\partial u}}{{\partial y}}\frac{{\partial y}}{{\partial r}},\quad \frac{{\partial u}}{{\partial \theta }} = \frac{{\partial u}}{{\partial \theta }}\frac{{\partial x}}{{\partial \theta }} + \frac{{\partial u}}{{\partial y}}\frac{{\partial y}}{{\partial \theta }}.\)
Or that can be written as:
\(\displaystyle \L \left\{ \begin{array}{l}
u_r = u_x \cos (\theta ) + u_y \sin (\theta ),\quad u_\theta = - u_x r\sin (\theta ) + u_y r\cos (\theta ) \\
\mbox{likewise} \\
v_r = v_x \cos (\theta ) + v_y \sin (\theta ),\quad u_\theta = - v_x r\sin (\theta ) + v_y r\cos (\theta ) \\
\end{array} \right.\)
Because f is differentiable with continuous partial derivatives, using the Cauchy-Riemann equations we get: \(\displaystyle u_x = v_y \quad \& \quad u_y = - v_x .\)
Solving the above (you can do that messy algebra, I won’t) we get:
\(\displaystyle \L u_r = \frac{{v_r }}{r}\quad \& \quad v_r = \frac{{ - u_\theta }}{r}.\)