How would one go about showing the polar version of the Cauchy Riemann Equations are sufficient to get differentiability of a complex valued function which has continuous partial derivatives?
I haven't found any proof of this online.
One of my ideas was writing out $r$ and $\theta$ in terms of $x$ and $y$, then taking the partial derivatives with respect to $x$ and $y$ and showing the Cauchy Riemann equations in the Cartesian coordinate system are satisfied. A problem with this approach is that derivatives get messy.
What are some other ways to do it?
Best Answer
I happen to have some notes on this question. What follows here is the usual approach, it's just multivariate calculus paired with the Cauchy Riemann equations. I have an idea for an easier way, I'll post it as a second answer in a bit if it works.
If we use polar coordinates to rewrite $f$ as follows: $$ f(x(r,\theta),y(r,\theta)) = u(x(r,\theta),y(r,\theta))+iv(x(r,\theta),y(r,\theta)) $$ we use shorthands $F(r,\theta)=f(x(r,\theta),y(r,\theta))$ and $U(r,\theta )=u(x(r,\theta),y(r,\theta))$ and $V(r,\theta )=v(x(r,\theta),y(r,\theta))$. We derive the CR-equations in polar coordinates via the chain rule from multivariate calculus, $$ U_r = x_ru_x + y_ru_y = \cos(\theta)u_x + \sin(\theta)u_y \ \ \text{and} \ \ U_{\theta} = x_{\theta}u_x + y_{\theta}u_y = -r\sin(\theta)u_x + r\cos(\theta)u_y $$ Likewise, $$ V_r = x_rv_x + y_rv_y = \cos(\theta)v_x + \sin(\theta)v_y \ \ \text{and} \ \ V_{\theta} = x_{\theta}v_x + y_{\theta}v_y = -r\sin(\theta)v_x + r\cos(\theta)v_y $$ We can write these in matrix notation as follows: $$ \left[ \begin{array}{l} U_r \\ U_{\theta} \end{array} \right] = \left[ \begin{array}{ll} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{array} \right]\left[ \begin{array}{l} u_x \\ u_y \end{array} \right] \ \ \text{and} \ \ \left[ \begin{array}{l} V_r \\ V_{\theta} \end{array} \right] = \left[ \begin{array}{ll} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{array} \right]\left[ \begin{array}{l} v_x \\ v_y \end{array} \right] $$ Multiply these by the inverse matrix: $\left[ \begin{array}{ll} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{array} \right]^{-1} = \frac{1}{r}\left[ \begin{array}{ll} r\cos(\theta) & -\sin(\theta) \\ r\sin(\theta) & \cos(\theta) \end{array} \right]$ to find $$ \left[ \begin{array}{l} u_x \\ u_y \end{array} \right] = \frac{1}{r}\left[ \begin{array}{ll} r\cos(\theta) & -\sin(\theta) \\ r\sin(\theta) & \cos(\theta) \end{array} \right]\left[ \begin{array}{l} U_r \\ U_{\theta} \end{array} \right] = \left[ \begin{array}{l} \cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta} \\ \sin(\theta)U_r + \tfrac{1}{r}\cos(\theta)U_{\theta} \end{array} \right] $$ A similar calculation holds for $V$. To summarize: $$ u_x = \cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta} \ \ \ \ v_x = \cos(\theta)V_r - \tfrac{1}{r}\sin(\theta)V_{\theta} $$ $$ u_y =\sin(\theta)U_r + \tfrac{1}{r}\cos(\theta)U_{\theta} \ \ \ \ v_y =\sin(\theta)V_r + \tfrac{1}{r}\cos(\theta)V_{\theta} $$ The CR-equation $u_x=v_y$ yields: $$ (A.) \ \ \cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta} = \sin(\theta)V_r + \tfrac{1}{r}\cos(\theta)V_{\theta} $$ Likewise the CR-equation $u_y=-v_x$ yields: $$ (B.) \ \ \sin(\theta)U_r + \tfrac{1}{r}\cos(\theta)U_{\theta} = -\cos(\theta)V_r + \tfrac{1}{r}\sin(\theta)V_{\theta}$$ Multiply (A.) by $r\sin(\theta)$ and $(B.)$ by $r\cos(\theta)$ and subtract (A.) from (B.): $$ \boxed{U_{\theta} = -rV_r} $$ Likewise multiply (A.) by $r\cos(\theta)$ and $(B.)$ by $r\sin(\theta)$ and add (A.) and (B.): $$ \boxed{rU_r = V_{\theta}} $$ Finally, recall that $z = re^{i\theta}=r(\cos(\theta)+i\sin(\theta))$ hence \begin{align} \notag f'(z) &= u_x+iv_x \\ \notag &= (\cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta})+i(\cos(\theta)V_r - \tfrac{1}{r}\sin(\theta)V_{\theta}) \\ \notag &= (\cos(\theta)U_r + \sin(\theta)V_{r})+i(\cos(\theta)V_r - \sin(\theta)U_{r}) \\ \notag &= (\cos(\theta)- i\sin(\theta))U_r + i(\cos(\theta)-i\sin(\theta))V_r \\ \notag &= e^{-i\theta}( U_r+iV_r) \notag \end{align}