Let $f(z)$ be a rational function in the complex plane such that $f$ does not have any poles in $\{z:\Im z\ge0\}$.
Prove that $\sup\{|f(z)|:\Im z\ge0\}=\sup\{|f(z)|:\Im z=0\}$.
Let $\Gamma_r$ be a half circle counter such that $\Gamma_r=\Gamma_{r_1}+\Gamma_{r_2}$ when $\Gamma_{r_1}=\{z:\Im z=0, |z|=r\}$,$\Gamma_{r_2}=\{z:\Im z>0, |z|=r\}$. Using the Maximum modulus principle on the insides of $\Gamma_r$, $|f|$ Gets is maximum value on $\Gamma_r$. As $r$ gets bigger if $|f|$ got it's maximum on $\Gamma_{r_2}$ than it's still smaller the the value on $\Gamma_{r+1}$ which does not contain $\Gamma_{r_2}$. I would like a hint on how to proceed.
To develop the comment of Daniel Fischer. Let $T(z)=\dfrac{i-z}{1-iz}$. Then $T$ is a conformal mapping of $D(0,1)$ onto the upper half plane $\{z:\Im z> 0\}$. $f\circ T$ is a rational function which is analytic in $D(0,1)$ and continuous on $\overline{D(0,1)}$, according to the assumption. So the maximum of $|f\circ T|$ on $\overline{D(0,1)}$ is attained on the boundary of the disk that is $$ \sup_{D(0,1)}|f\circ T|=\sup_{C(0,1)}|f\circ T| $$ or equivalently $$ \sup_{\{z:\Im z>0\}}|f(z)|=\sup_{\{z:\Im z=0\}}|f(z)|. $$ Because $T(D(0,1))=\{z:\Im z>0\}$, and $T(C(0,1)\setminus\{-i\})=\{z:\Im z=0\}$.