Let $G$ be a region in a complex plane. Suppose $f:G\to\mathbb C$ is analytic and one-one. Show that $f'(z)$ is not zero for any $z$ in $G$.
I was trying to do the contradiction method. But I can't conclude anything. Please help
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Cave Johnson
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@CaveJohnson What's the flaw? – José Carlos Santos Apr 11 '18 at 13:25
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1@JoséCarlosSantos There are two flaws. First, $n\ge2$ does not mean that we may write $f(z)=(z-z_0)^2g_1(z)^2$. This only works for $n=2$. Second, I see no reason that $f(w)$ should be equal to $f(w')$. For example, take $f(z)=z^2\exp(z)$, then $f(1)=1$ while $f(-1)=e^{-1}$. – Cave Johnson Apr 11 '18 at 13:28
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@CaveJohnson You are right about both flaws. However, the first one is nor important, since we're only interested in the case $n=2$ here. On the other hand, the second flaw is important. I will retract my closing vote and, if the sysem allows me (I'me not sure about that), I will support your suggestion. – José Carlos Santos Apr 11 '18 at 13:48