I think I lack intution when dealing with disproving holomorphicity of functions like:
a) $f(z) = \frac{z}{\lvert z \rvert}$
b) $f(z) = \left \{ \begin{array} .\frac{z^5}{\lvert z^4 \rvert} &z\neq 0\\0 & z=0 \end{array} \right . $
For a), one way to prove it is that we have constant modulus, therefore $f$ can't be holomorphic since in that case it would be constant (using the Open Mapping Theorem or the Cauchy Riemann equations). Is there an alternative way of proving this conveniently?
For b), I am unsure, since we don't have constant modulus. We can write $z = |z|e^{i \varphi}$ using polar coordinates and get $$f(z) = \left \{ \begin{array} . \lvert z \rvert e^{5i \varphi} &z\neq 0\\0 & z=0 \end{array} \right .$$ This to me looks similar like the identity, besides the $5$ in the exponent. How do I prove this one isn't holomorphic (at $z=0$)?
In mathematics, it is often useful to simply go back to the definition. A function is complex differentiable if for all $z_0 \in \mathbb{C}$, $\frac{f(z)-f(z_0)}{z-z_0}$ has a limit when $z \to z_0$.
Here there is clearly a problem at $z_0 = 0$. So let's try to compute $\lim_{z \to 0} f(z)/z$.
For a), $\lim_{z \to 0} f(z)/z = (z/|z|)/z = 1/|z|$ which does not converge to anything as $t \to 0$. So the limit doesn't exist.
For b), it's a bit more subtle. A useful property in complex analysis is that if a limit exists, then it must be the same limit along all path leading to the point $z_0$. Let's try to compute the limit along a horizontal vertical line, i.e. with $z = t$, $t \in \mathbb{R}$ and $t \to 0$ (distinguishing the two cases $t>0$ and $t<0$).
For the first one, $f(t) = t^5/|t|^4$. If $t > 0$, this gives $f(t) = t$ and so $f(t)/t = 1$ and the limit is $1$. But if $t < 0$, then $f(t) = -t$ and $f(t)/t = -1$ and the limit is $-1$. But to be complex differentiable, the limit should have been the same, a contradiction. So the function is not complex differentiable.