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$f(z)$ is a holomorphic function on $B$, where $B=D\cup \{1\}, D=\{z: |z|<1\}$.
$f(D)\subset D, f(1)=1$. prove that $f'(1)\geq 0$

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Since $1$ is not in the interior of $B$, how do you define $f$ holomorphic on $B$. In particular, what does $f'(1)$ mean? –  1015 Feb 4 '13 at 0:54
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It is customary to also include your thoughts on the problem. –  Antonio Vargas Feb 4 '13 at 0:54
    
Extending julien's question, do you mean that $f$ is holomorphic on a neighborhood of $B$, or is $f'(1)$ defined as $\lim\limits_{z\to 1,z\in D}\dfrac{f(z)-f(1)}{z-1}$? –  Jonas Meyer Feb 4 '13 at 0:55
    
@julien: The standard usage when blah is not open is to say that $f$ is holomorphic on blah iff for every point $x$ of blah there is an open neighborhood $V$ of $x$ and a holomorphic function $g$ on $V$ that coincides with $f$ on $V\cap$blah. Here, this just means that there is an open set containing $D\cup B$ where $f$ is holomorphic. –  Andres Caicedo Feb 4 '13 at 0:57
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Hint: if $f'(1) = |f'(1)| e^{i\theta}$ for $\theta \ne 0$, say $\theta > 0$, what will happen to the image of the curve $t \mapsto 1+ t e^{i\phi}$ under $f$ when $\phi \in (\frac{\pi}{2}-\theta,\frac{\pi}{2})$. –  achille hui Feb 4 '13 at 1:35

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