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Let $D=\{z:|z|<1\}$. suppose $f(z)$ is analytic on $D$, $\mathrm{Re}f(z)>0,\forall z\in D, f(0)>0$. show that

i) $$|\mathrm{Im}f(z)|\leq f(0)\frac{2|z|}{1-{|z|^2}}$$ ii) $$f(0)\frac{1-|z|}{1+|z|}\leq \mathrm{Re}f(z)\leq |f(z)| \leq f(0)\frac{1+|z|}{1-|z|}$$ "=" holds if and only if $f(z)=w_0\dfrac{1+e^{i\alpha}z}{1-e^{i\alpha}z}$($w_0,\alpha\in\Bbb R$,and $w_0>0$)

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1 Answer 1

Hints:
1) Show that for all $z\in D$ $$\frac{1}{2\pi}\int_0^{2\pi}\operatorname{Re}(f(e^{it}))\frac{e^{it}+z}{e^{it}-z}dt\hspace{3pt}=\hspace{3pt}f(z)-i\operatorname{Im}(f(0))$$ This can be done by substituting $\operatorname{Re}(f(e^{it}))=\frac{f(e^{it})+\overline{f(e^{it})}}{2}$ and calculating using Cauchy's integral theorem.
2) Apply absolute value on both sides, use integral triangle inequality and the fact that $\operatorname{Re}(f(z))>0$ and that given $f(0)>0$ implies $\operatorname{Im}(f(0))=0$.

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Thanks! How to get $|\mathrm{Im}f(z)|\leq f(0)\frac{2|z|}{1-{|z|^2}}$ ? I got 2) –  ziang chen Feb 6 '13 at 22:32

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