Let $f(z)$ be a holomorphic function and $f(z)$ is 1-1.
If $f(z)$ has a zero at $z=x$,
then $x$ is a simple zero?
I guess this is simple fact, But I can't prove it, and I can't find related article.
Thanks!
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@MartinR Why? I can not understand.. – nicksohn Dec 30 '15 at 11:43
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$f$ has a simple zero at $z$ if $f'(z) \ne 0$, that's how a simple zero is defined. – Martin R Dec 30 '15 at 11:45
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But non-zero means it is not identically 0, So it can be 0 at z.. am i wrong? – nicksohn Dec 30 '15 at 11:49
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No. The referenced thread shows that if $f$ is one-to-one then $f'(z) \ne 0$ for all $z$ in the domain. – Martin R Dec 30 '15 at 11:50
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Ah I got it. Thanks. But it looks overkill... isn't it? :) – nicksohn Dec 30 '15 at 11:50
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If $f'(z_0)=\cdots=f^{(k-1)}(z_0)=0$ and $f^{(k)}(z_0)\ne 0$., then there is an open neighbourhood $U$ of $z_0$, such that $f$ takes in $U\setminus\{z_0\}$ every value exactly $k$ times.
Yiorgos S. Smyrlis
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Sorry I couldn't follow this. Could you tell me more specific explanation? – nicksohn Dec 30 '15 at 11:13