We have a statement “If $f(z)$ is not an entire function, then $g(z) = f^2(z)$ cannot be an entire function.”.
Two students and a professor had the following conversation:
A: The statement is false. I can give a counter-example. Let $f(z) = \sqrt z$, assuming that it is a branch that takes −1 to i. Then it is not analytic on the branch cut, but $g(z) = f ^2(z) = z$ is obviously an entire function.
B: I think this is not a counter-example, so we cannot conclude the statement is false yet.
Professor: Good! B is correct. The statement is false.
Now I give you an assignment:
(a) Show that student A’s example does not work as a counter-example.
Why doesn't student A's example work as a counter example? I don't understand what is wrong with the statement that it'd be incorrect
Also I'm not exactly clear on what $f^2(z)$ means is it $f(f(z))$ or $[f(z)]^2$
This is a bit different from my previous question as previously I was asking about why it wasn't analytic on the branch cut, now I'm trying to understand why the counterexample: $f(z)=\sqrt z$ isn't right. I know $f(z)$ is not analytic on the branch -1 to i therefore not entire, but $f^2(z)=z$ is clearly analytic, and therefore entire so it should work as a counterexample however the question states it doesn't. I don't understand why it doesn't work as a counter example. No matter how I look at it $z$ is obviously entire and $\sqrt z$ isn't so shouldn't it work as a proper counter example.