Question

Let f(z)=1/z^5 . Use the polar form of the Cauchy Riemann equations to determine where f is differentiable

Answer 1

$z=re^{i\theta},$ where $r>0$ and $\theta\in[0,2\pi)$. This is the polar form of the complex variable. $f=r^{-5}e^{-5i\theta}.$ Remind that $e^{i\theta}=cos(\theta)+i\,sin(\theta)$. We receive: $f=r^{-5}(cos(5\theta)-i\,sin(5\theta))=u+iv,$ where $u=r^{-5}cos(5\theta)$, $v=-r^{-5}sin(5\theta)$. Cauchy-Riemann equations in polar coordinates have the following form: $\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial\theta}$, $\frac{\partial v}{\partial r}=-\frac1r\frac{\partial u}{\partial\theta}$. $\frac{\partial u}{\partial r}=-5r^{-6}cos(5\theta),$ $\frac{\partial u}{\partial \theta}=-5r^{-5}sin(5\theta)$, $\frac{\partial v}{\partial r}=5r^{-6}sin(5\theta),$ $\frac{\partial v}{\partial\theta}=-5r^{-5}cos(5\theta)$. As we can see, the Cauchy-Riemann equations are satisfied at all points, except $r=0$ . Functions $u,v$ are continuously differentiable at all points, except $r=0$. Thus, due to the respective Theorem from complex analysis, the function $f$ is differentiable at all points except .