Question about holomorphic functions and zero's in complex analysis

Question

The question: Write an example of a connected, open set $U ⊂ C$ and an analytic function $f : U → C$ such that $f(z)$ has $N$ zeros on $U$ while $f′(z)$ never vanishes on $U$

My guess is that I was going to write a function something like this:

$f(z) = (z-z_1)(z-z_2)*...*(z-z_N)exp(\frac{z}{z_1} + z/z_2 + ... + z/z_N )$ However I think the derivative would not vanish. Am I on the right track with this example. Thank you very much!

Answer 1

I don't see why $f'(z)$ would never be $0$, for your function $f$.

On the other hand, you can take $g(z)=e^z-1$, with $U$ open, connected and such that $0,2\pi i,4\pi i,\ldots,2(N-1)\pi i\in U$.

Answer 2

Take whatever analytic function $f$ you want, as long as it has at least $N$ zeros which are not zeros of $f'$, and you can choose your $U$ to include those $N$ zeros of $f$ and no zeros of $f'$. Your function would work fine, except that it's going to be hard to find the zeros of $f'$ explicitly. Somewhat simpler would be $$f(z) = (z-1) \ldots (z - N)$$ where we know all the zeros of $f'$ are real. Then let $U$ consist of, say, the open upper half plane together with some small open disks around $1, \ldots, N$, small enough to not contain any zeros of $f'$.

Answer 3

Let $f(z)=z^N-1$ and take $U$ as the open annulus of large radius $2$ and small radius $1/2$: