Prove/Provide counter-example: complex function is a. not entire b. continuous

Question

Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be an entire complex function that is non-constant. We define $g : \mathbb{C} \rightarrow \mathbb{C}$ as follows:

$g(z)= \begin{cases} 1&\text{if}\, f(z)=0\\ \frac{|f(z)|+1}{|f(z)|}.f(z) &\text{if}\, f(z)\neq0\ \end{cases}$

Prove/ provide a counter example that the function $g(z)$ is a. not entire b. continuous in $\mathbb{C}$.

Since $f$ is entire and non-constant I conluded that $f$ is not bounded from Liouville's theorem, I believe $g$ is not entire and non-continuous. How do I go about proving both? Thank you.

Answer 1

a) It cannot be entire. Hint. Show that these functions are not complex-differentiable at points where $f(z) \neq 0$: $$ h_1(z) = |f(z)|, \quad h_2(z) = \frac{|f(z)|+1}{|f(z)|}, \quad h_3(z) = \frac{|f(z)|+1}{|f(z)|} f(z). $$ You may want to start with $h_1$ (open mapping theorem should help, although there are easier ways), then express $h_1$ in terms of $h_2$ and $h_2$ in terms of $h_3$.

b) The function $f(z) = z$ is a counterexample.

Answer 2

If $f(z)=z$, then you can check easily that $$\lim_{\alpha \rightarrow 0 \\ \alpha >0}g(i\alpha) = i \neq 1 =g(0)$$

so $g$ is not continuous.