Let $f(z)$ be a rational function in the complex plane such that $f$ does not have any poles in $\{z:\mathrm{\Im}z\ge 0\}$.

Prove that $sup\{|f(z)|:\mathrm{\Im}z\ge 0\}=sup\{|f(z)|:\mathrm{\Im}z=0\}$.

Let ${\mathrm{\Gamma}}_{r}$ be a half circle counter such that ${\mathrm{\Gamma}}_{r}={\mathrm{\Gamma}}_{{r}_{1}}+{\mathrm{\Gamma}}_{{r}_{2}}$ when ${\mathrm{\Gamma}}_{{r}_{1}}=\{z:\mathrm{\Im}z=0,|z|=r\}$, ${\mathrm{\Gamma}}_{{r}_{2}}=\{z:\mathrm{\Im}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma}}_{r}$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma}}_{r}$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma}}_{{r}_{2}}$ than it's still smaller the the value on ${\mathrm{\Gamma}}_{r+1}$ which does not contain ${\mathrm{\Gamma}}_{{r}_{2}}$. I would like a hint on how to proceed.

Prove that $sup\{|f(z)|:\mathrm{\Im}z\ge 0\}=sup\{|f(z)|:\mathrm{\Im}z=0\}$.

Let ${\mathrm{\Gamma}}_{r}$ be a half circle counter such that ${\mathrm{\Gamma}}_{r}={\mathrm{\Gamma}}_{{r}_{1}}+{\mathrm{\Gamma}}_{{r}_{2}}$ when ${\mathrm{\Gamma}}_{{r}_{1}}=\{z:\mathrm{\Im}z=0,|z|=r\}$, ${\mathrm{\Gamma}}_{{r}_{2}}=\{z:\mathrm{\Im}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma}}_{r}$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma}}_{r}$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma}}_{{r}_{2}}$ than it's still smaller the the value on ${\mathrm{\Gamma}}_{r+1}$ which does not contain ${\mathrm{\Gamma}}_{{r}_{2}}$. I would like a hint on how to proceed.