An inequality involving a holomorphic function Suppose $f \in H(\mathbb{D})$ and $f(0)=0 , |Ref(z)|<1$ . Then prove that : $$|f(z)|\le \frac{2}{\pi}\log\left(\frac{1+r}{1-r}\right)$$ for $|z|\le r<1$ and $$|Ref(z)|\le \frac{4}{\pi}\tan^{-1}(|z|)$$. I tried showing this by using the fact the unit disc is a convex domain so it retains the hyperbolic metric. But i think that it can be solved by using that f takes the unit disc to the strip $|Ref|<1$
 A: Let $D_r=\{z : |z|\le r\}$ and $\varphi (z)=\frac{2}{\pi i}\log \left(\frac{1+z}{1-z}\right)$, which maps $\mathbb{D}$ to the strip $|\operatorname{Re}\, z|<1$ bijectively. 
First we prove 
\begin{align}
\left|\log \left(\frac{1+z}{1-z}\right)\right|\le \log \left(\frac{1+r}{1-r}\right) \quad (|z|\le r<1).\tag{1}
\end{align}
In other words
$$
|w|\le \frac{2}{\pi}\log \left(\frac{1+r}{1-r}\right)
$$
holds for all $w\in \varphi (D_r)$.
Proof.\begin{align}
\log \left(\frac{1+z}{1-z}\right)&=\log(1+z)-\log(1-z)\\
&=\int_\gamma \frac{d\zeta }{1+\zeta }+\int_\gamma \frac{d\zeta }{1-\zeta } \quad (\gamma : \zeta =zt\,\,(0\le t\le 1))\\
&=\int_0^1 \frac{zdt}{1+zt}+\int_0^1 \frac{zdt}{1-zt}=2z\int_0^1 \frac{dt}{1-z^2t^2}.
\end{align}
Taking absolute values we have \begin{align}
\left|\log \left(\frac{1+z}{1-z}\right)\right|&\le 2|z|\int_0^1  \frac{dt}{1-|z|^2t^2}\\
&=\int_0^1 \frac{|z|dt}{1+|z|t}+\int_0^1 \frac{|z|dt}{1-|z|t}\\
&=\log(1+|z|)-\log(1-|z|)=\log\left(\frac{1+|z|}{1-|z|}\right)\\
&\le \log\left(\frac{1+r}{1-r}\right).
\end{align}
The proof is complete.
Now we consider $$
g(z)=(\varphi ^{-1}\circ f)(z).$$
Since $g$ satisfies $g(0)=0$ and $|g(z)|<1$, by Schwarz's lemma we see that $|g(z)|\le |z|,$  that is, $g(D_r)\subset D_r.$ 
This implies $$
f(D_r)=(\varphi \circ g)(D_r)\subset \varphi (D_r).$$
Therefore we see that$$
|f(z)|\le \frac{2}{\pi}\log \left(\frac{1+r}{1-r}\right)\quad (|z|\le r<1)$$
holds.
Next we prove that
$$
|\operatorname{Re}\, f(z)|\le \frac{4}{\pi}\tan^{-1} r$$
for $z$ with $|z|\le r$.
Let $\phi(z)=\frac{1+z}{1-z}.$ Then $\phi(D_r)$ is the disk centered at $z=\frac{1+r^2}{1-r^2}$ with radius $\frac{2r}{1-r^2}$. Therefore we see that $$
|\arg \phi(z)|\le \sin^{-1} \left(\frac{2r}{1+r^2}\right)=2\tan^{-1}r$$
for $z$ with $|z|\le r$. 
Thus we know that $$
|\operatorname{Re}\varphi (z)|\le \frac{4}{\pi}\tan^{-1}r  \quad (|z|\le r),
$$
which implies that $$
|\operatorname{Re}f(z)|\le \frac{4}{\pi}\tan^{-1}r  \quad (|z|\le r),$$
since $f(D_r)\subset \varphi (D_r).$
