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Exercise 3. Let $f$ be analytic in $\overline{B}(0; R)$ with $f(0)=0$, $f'(0) \neq 0$ and $f(z) \neq 0$ for $0<|z| \leq R$. Put $\rho=\min\{|f(z)|:|z|=R\}>0$. Define $g: B(0; \rho) \rightarrow \mathbb{C}$ by $$g(w) = \frac{1}{2\pi i} \int_\gamma \frac{z\,f'(z)}{f(z)-\omega} dz$$ where $\gamma$ is the circle $|z|=R$. Show that $g$ is analytic and discuss the properties of $g$.

Now, I want to apply Rouche's theorem in this question, so please i want some hints. I think $f$ maybe one to one since the kernel maybe the zero element.

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  • $\begingroup$ I think f maybe one to one since the kernel maybe the zero element $\endgroup$ – Abdallah May 7 '15 at 3:04
  • $\begingroup$ "I think $f$ maybe one to one since the kernel maybe the zero element" reflects a lot of confusion. (1) $f$ is not a linear transformation or homomorphism between groups, so what do you mean by "kernel"? (2) If you meant the set $\{z: f(z) = 0\}$, having that set equal $\{0\}$ does not imply that $f$ is injective. $\endgroup$ – Jonas Meyer May 27 '15 at 3:27
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I don't see any use for Rouché here. Just consider the poles of the integrand inside $\gamma$, and what their residues are.

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  • $\begingroup$ Sorry but g(w) must be the inverse of f and this is an application of rouche's theoram $\endgroup$ – Abdallah May 7 '15 at 3:07
  • $\begingroup$ Yes, I know that $g$ is the inverse of $f$. I guess you could use Rouché to show that $f$ is one-to-one (i.e. that the number of zeros of $f(z)-w$ is the same as the number of zeros of $f(z)$, which is $1$). Or you could use the Argument Principle. $\endgroup$ – Robert Israel May 7 '15 at 4:20

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