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Currently working on a head-scratcher:

Let $f: B_1(0) \rightarrow \mathbb{C}$ holomorphic, s.t.: For $z \in \mathbb{C}$ with $\vert z \vert = \frac{1}{2}$ it holds: $$\vert f'(z) - z\exp(z) \vert < \frac{1}{2}\exp(\Re(z))$$

Then f is not injective.

I'm not really sure how to approach this. Has anyone ever seen such a problem before and has any intuition on how to approach it? I just don't see how a statement about the derivative would imply something about the original function.

Zedssad
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1 Answers1

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For $|z| = 1/2$ is $$ \vert f'(z) - z\exp(z) \vert < \frac{1}{2}\exp(\Re(z)) = |z\exp(z) | $$ which implies by Rouché's theorem that $f'$ and $z\exp(z)$ have the same number of zeros in the disk $|z|< 1/2$.

So $f'$ has a zero in that disk, and therefore $f$ is not injective (see for example Proof that 1-1 analytic functions have nonzero derivative).

Martin R
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