$$f(z)=\left\{\begin{array}{cc}(z^5)/(|z|)^4 &\text{when}~z\neq0,\\0&\text{when}~z=0\end{array}\right.$$. For this function, how would I prove that $f(z)$ is not differentiable at $z=0$?
I tried a taking the limit of $(f(\delta z) - f(0))/(\delta z)$ as $\delta z\rightarrow0$, but the result seems to indicate that the function IS differentiable (which I know is not true).
$$f(\Delta z) / \Delta z = (\Delta z/|\Delta z|)^4$$ If you take $\Delta z=h \in \mathbb{R}$ real this is equal to $1$, but if you take $\Delta z = \omega h$ with $h \in \mathbb R$ and $\omega$ such that $\omega^4 = -1$, it's $-1$. Because there is points of each type arbitrarily close to the origin, the limit doesn't exist.
Let $z=re^{i\theta}$ so by definition we have that \begin{align} f'(0)&=\lim\limits_{z\to 0} \frac{f(z)-f(0)}{z-0}\\ &=\lim\limits_{z\to 0} \frac{z^5}{z|z|^4}\\ &=\lim\limits_{z\to 0} \frac{z^4}{|z|^4}\\ &=\lim\limits_{r\to 0} \frac{r^4e^{4i\theta}}{r^4}\\ &=e^{4i\theta} \end{align} This shows that $f(z)$ is not differentiable at $z=0$
