Polar form of Laplace's equation. 
Let the function $f(z)=u(r,\theta)+iv(r,\theta)$ be analytic in a domain $D$ that does not contain the origin. Using the Cauchy-Riemann equations in polar coordinates and assuming continuity of partial derivatives, show that throughout $D$ the function $u(r,\theta)$ satisfies the partial differential equation
  $$r^2u{_r}{_r}(r,\theta)+ru_r(r,\theta)+u{_\theta}{_\theta}=0$$ 
  which is the polar form of Laplace's equation. Show that the same is true of the function $v(r,\theta)$.
  I know $H{_x}{_x}(x,y)+H{_y}{_y}(x,y)=0$ but where to from here?

 A: Take the Cauchy-Riemann equations in polar form, that is $$u_r = \frac{1}{r}v_{\theta},\quad \frac{1}{r} u_{\theta} = -v_r .$$ Now, take the partial derivative of the first equation with respect to $r$ and the partial derivative of the second equation with respect to $\theta$, then $$ u_{rr} = \frac{\partial}{\partial r} (\frac{1}{r}v_{\theta}) = \frac{v_{\theta\ r}}{r} - \frac{v_{\theta}}{r^2},\\ u_{\theta \theta} = \frac{\partial}{\partial \theta}(-r v_r)= - r v_{r \theta} .$$ Here, I just took the partial derivative for the first one with respect to $r$ and the partial derivative with respect to $\theta$ for the second one, but you can work out the other two partial derivatives. Now, assuming continuity of partial derivatives, one can say that $v_{r \theta} = v_{\theta r},$ so we can substitute the last equation on the one before, thus $$u_{rr} = -\frac{u_{\theta \theta}}{r^2} - \frac{v_{\theta}}{r^2},$$ but we can again make use of the Cauchy-Riemann equations and substitue $v_{\theta} = r u_r,$ and using this result in the last equation yields $$ r^2 u_{rr} = -u_{\theta \theta} - r u_r,\\r^2 u_{rr} + r u_r + u_{\theta \theta} = 0.$$ 
