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Let $f$ be a holomorphic function from $H=\{z \in \mathbb C : \operatorname{Im}(z)>0\}$ to $D=\{z \in \mathbb C : |z|<1\}$. Suppose that $f(z_0)=f'(z_0)=0$.

Show that: \[ |f(z)| \leq \frac{|z-z_0|^2}{|z-\overline{z}_0|^2} \quad\text{and}\quad |f''(z_0)| \leq \frac{1}{2|\operatorname{Im}(z_0)|^2} \]

Thanks everyone for your help!!

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What have you tried? In my experience, questions that show some effort beyond a reproduction of an exercise get more attention. If you've tried something and it didn't work, it's okay and even desirable to mention that. My gut reaction is to use some conformal mapping to get a map $D \to D$ and then apply Schwarz or Shwarz–Pick, but I don't have time to pursue this at the moment. Hopefully someone else will. Welcome, by the way! – Dylan Moreland Jun 27 '12 at 18:35
Hey Dylan! First of all thanks for the greeting! I just tried to call $g(z)=f(z)/|z-z_0|^2$ and using the maximum modulus principle as when you demonstrate that $|f(z)|<|z|$ when f goes from D to worked for demonstrating that $|f(z)|<|z|^2$, but I can't solve this one... – Nicolo Canali De Rossi Jun 27 '12 at 18:41
Should the $z$ in your second inequality be $z_0$? – Robert Israel Jun 27 '12 at 19:02
yep, I mistyped – Nicolo Canali De Rossi Jun 27 '12 at 19:35
And by the way solving the first inequality the second becomes pretty obvious to me...the problem is the first one! – Nicolo Canali De Rossi Jun 27 '12 at 19:37

Hint: let $g = f \circ \varphi^{-1}$ where $\varphi(z) = \dfrac{z - z_0}{z - \overline{z_0}}$, and note that $g(w)/w^2$ has a removable singularity at $0$.

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