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Let $f$ be analytic on the unit disk $D = \{z : |z| ≤ 1\}$ and suppose $Im(f(z)) > 0$ for $z ∈ D$ and $f(0) = i$. Prove that $|f ' (0)| ≤ 1$. For what functions do we have equality?

I'm not sure how to go about this. Any solutions or hints are greatly appreciated.

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closed as off-topic by heropup, JKnecht, user147263, Daniel W. Farlow, user296602 Apr 22 '16 at 4:45

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  • $\begingroup$ Since January 20 all but one of the questions you have posted have lacked context. Several of your posts have been closed for this reason, yet you continue to ignore all warnings that the posting of questions without any explanation of your own thoughts or effort to solve them is not allowed on this site. $\endgroup$ – heropup Apr 21 '16 at 20:21
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The statement is false. Set $$ f(z)=i\frac{1-rz}{1+rz} $$ defined on the disk $D_{1/r}:= \{ z: |z|<1/r\}$ where $r$ is chosen to be $0<r<1$ and $1/r$ close enough to $1$. Then $f$ is holomorphic on $D_{1/r}$, hence on $D$. It maps $D_{1/r}$ onto the upper half plane, $f(0)=i$ but $f'(0)=-2ri$ so that $$ |f'(0)|=2r>1 $$ if $r$ was taken to be greater than $1/2$.

However, if $f$ is holomorphic on $\mathbb D:= \{ z: |z|<1\}$, $\text{Im} (f(z))>0$ for all $z\in \mathbb D$ and $f(0)=i$, then one can deduce that $|f'(0)|\le 2$. The proof follows from the

Schwarz lemma: If a holomorphic function $f:\mathbb D\to \mathbb D$ satisfies $f(0)=0$, then $|f(z)|\le |z|$ on $\mathbb D$ and $|f'(0)|\le 1$.

The proof of this theorem can be found in any textbook so I omit it. Let $h:=g\circ f$ where

$$ g(z)=\frac{z-i}{z+i}. $$ Then $g$ is holomorphic and it maps upper half plane onto $\mathbb D$. Hence $h$ is holomorphic, $h(0)=g(i)=0$ and $|h(z)|<1$ on $\mathbb D$. Applying Schwarz's lemma to $h$, we conclude that

$$ |h'(z)|=\left| \frac{2i}{(f(z)+i)^2}\cdot f'(z)\right|\le 1. $$ Consequently, $|f'(0)|\le |2i|=2$.

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Hint: Let $g(w) = (w-i)/(w+i)$ Then $g$ maps the open upper half plane into $\mathbb D$ and $g(i)=0.$ Consider $g\circ f.$

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  • $\begingroup$ Why do I consider that composition? $\endgroup$ – Happy Apr 18 '16 at 3:10
  • $\begingroup$ Because it's a holomorphic map from the disc to the disc that sends 0 to 0. That should ring bells $\endgroup$ – zhw. Apr 18 '16 at 3:52
  • $\begingroup$ all right, who downvoted my answer. it was a good hint $\endgroup$ – zhw. Apr 21 '16 at 22:02

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