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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an holomorphic function such that $f'(z_0) \neq 0$ for some $z_0 \in \mathbb{C}$.Prove that there is $r>0$ such that, if $|z-z_0|<r$ and $z \neq z_0 $, then $f(z) \neq f(z_0)$.

It seems very trivial, but I can't solve it...

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up vote 4 down vote accepted

Write $$f(z) = f(z_0) + \varphi(z)(z-z_0)$$ where $\varphi: \mathbb C \to \mathbb C$ is continuous with $\varphi(z_0) = f'(z_0).$ By continuity of $\varphi$ at $z_0$, there is $r > 0$ such that $\varphi(z) \neq 0$ for all $z \in \mathbb C$ with $|z-z_0|< r$. Then, for those $z$, if $z \neq z_0$, $$|f(z)-f(z_0)| = |\varphi(z)||z-z_0| \neq 0,$$ so $f(z) \neq f(z_0)$.

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Since $f(z)$ is an holomorphic function, then the function $\tilde{f}(z)=f(z)-f(z_0)$ is holomorhic too with isolated zeroes and from Taylor's representation $$\tilde{f}(z)=f(z)-f(z_0)=\frac{f'(z_0)}{1!}(z-z_0)+o(z-z_0)\ne{0}.$$

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