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let $ B(0,1) = \{ z\in \mathbb{C} | |z|<1\} $ and $ f $ be an holomorphic function on $ B(0,1) $ such that $ f(z)\in\mathbb{R} \iff z\in\mathbb{R} $

Prove: $ f $ has at most 1 root in $ B(0,1) $

i think this exercise requires rouche theorem or the argument principle theorem but i cant see how to use it

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Let $g:(-1,1)\to\mathbb R$ be the restriction of $f$. All zeros of $f$ are zeros of $g$. If $g$ has two distinct zeros, then $g'(c)=0$ for some $c\in(-1,1)$ by Rolle's theorem. This implies that $f'(c)=0$, and that $h(z)=f(z)-f(c)$ has a zero of order at least $2$ at $c$. The argument principle implies that $h(z)$ takes on real values at least $4$ times as you go around a small enough circle centered at $c$, and only $2$ of these have real inputs.

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Would anyone care to elaborate on the last sentence ? How does the argument principle imply that $h(z)$ assumes real values at least 4 times ? –  Teddy Aug 1 '12 at 7:56

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