# Let f ( z ) be a rational function in the complex plane such that f does not ha

Let $f\left(z\right)$ be a rational function in the complex plane such that $f$ does not have any poles in $\left\{z:\mathrm{\Im }z\ge 0\right\}$.
Prove that $sup\left\{|f\left(z\right)|:\mathrm{\Im }z\ge 0\right\}=sup\left\{|f\left(z\right)|:\mathrm{\Im }z=0\right\}$.
Let ${\mathrm{\Gamma }}_{r}$ be a half circle counter such that ${\mathrm{\Gamma }}_{r}={\mathrm{\Gamma }}_{{r}_{1}}+{\mathrm{\Gamma }}_{{r}_{2}}$ when ${\mathrm{\Gamma }}_{{r}_{1}}=\left\{z:\mathrm{\Im }z=0,|z|=r\right\}$, ${\mathrm{\Gamma }}_{{r}_{2}}=\left\{z:\mathrm{\Im }z>0,|z|=r\right\}$. 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.
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potamanixv
To develop the comment of Daniel Fischer. Let $T\left(z\right)=\frac{i-z}{1-iz}$. Then T is a conformal mapping of D(0,1) onto the upper half plane $\left\{z:\mathrm{\Im }z>0\right\}$. $f\circ T$ is a rational function which is analytic in D(0,1) and continuous on $\overline{D\left(0,1\right)}$, according to the assumption. So the maximum of $|f\circ T|$ on $\overline{D\left(0,1\right)}$ is attained on the boundary of the disk that is
$\underset{D\left(0,1\right)}{sup}|f\circ T|=\underset{C\left(0,1\right)}{sup}|f\circ T|$
or equivalently
$\underset{\left\{z:\mathrm{\Im }z>0\right\}}{sup}|f\left(z\right)|=\underset{\left\{z:\mathrm{\Im }z=0\right\}}{sup}|f\left(z\right)|.$
Because $T\left(D\left(0,1\right)\right)=\left\{z:\mathrm{\Im }z>0\right\}$, and $T\left(C\left(0,1\right)\setminus \left\{-i\right\}\right)=\left\{z:\mathrm{\Im }z=0\right\}$.