I am not able to solve the following question and so I am asking for help here.
Theorem 11.2 : If f(z) is analytic at $z_0$ with $f(z_0) \neq 0$ , then f(z) is 1-1 in some nbd of $z_0$.
Question : Given a complex number $z_0$ and an $\epsilon >0$ , show that there exists a function f(z) analytic at $z_0$, $f'(z_0)\neq 0$ and such that f(z) is not 1-1 for $|z-z_0|< \epsilon$ . Does this contradicts Theorem 11.2?
I am unable to prove the result given in the question (independently) and also I think it will contradict the theorem 11.2 .
Let the complex number be $z_0$ and $f'(z_0)=a $, $a\neq 0$ and given that f(z) is analytic. But How to prove f(z) is not 1-1.
The sizes and shapes of nbd's in both cases can be different but both will have $z_0$ as centre.
So, I am not able to prove what is asked.
Can u please help me ?
Given $z_0 \in \Bbb C$ and $\epsilon > 0$, the function $$ f(z) = (z-z_0)(z-z_0+\frac 12 \epsilon) $$ is holomorphic with $f'(z_0) = \epsilon/2 \ne 0$. $f$ is not 1-1 on the disk $|z -z_0| < \epsilon$, since $f(z_0) = f(z_0-\epsilon/2)$.
This does not contradict Theorem 11.2. $f$ is still 1-1 in some smaller neighborhood of $z_0$, i.e. in some disk $|z -z_0| < \epsilon'$ with $0<\epsilon' < \epsilon$.