Finding An Entire Function If There Is 
Find if there is an entire function such that  $|f(x+iy)|=5e^{xy}$ and $f(i)=5$

I have tried to build system of equations to find $x,y$ 
from   $|f(x+iy)|=5e^{xy}$ we get $x^2+y^2=25e^{2xy}$ but this has led me to nowhere
 A: Observe that the function is never zero. Therefore, you can write $f(z)=e^{g(z)}$ for $g$ entire.
Therefore, $|f|=e^{\ln(5)+xy}$ implies that $Re(g)=\ln(5)+xy$. Using Cauchy Riemann you get that $D_y Im(g)=y$. Therefore $Im(g)=y^2/2+h(x)$ and that $h'(x)=-x$. Therefore $h(x)=-x^2/2+C$.
So, $g(z)=\ln(5)+xy -i(x^2-y^2)/2+Ci$. 
Since $f(i)=5$, we get that $\ln(5)=g(i)=i/2+Ci$. Therefore, $C=-i\ln(5)-1/2$.
Putting this back into $f$, you get $f(z)=5e^{-i\frac{z^2+1}{2}}$.
A: I recommend @fishroe's brilliant answer. And here is an alternative if you want.
Provided that
\begin{align}
x&=\frac{z+\bar{z}}{2},\\
y&=\frac{z-\bar{z}}{2i},
\end{align}
the original equation is equivalent to
$$
\left|f(z)\right|=5\exp\left(\frac{z+\bar{z}}{2}\frac{z-\bar{z}}{2i}\right)=5\exp\left(\frac{z^2-\bar{z}^2}{4i}\right),
$$
or equivalent to
$$
f(z)\overline{f(z)}=\left|f(z)\right|^2=25\exp\left(\frac{z^2-\bar{z}^2}{2i}\right)=\left(5\exp\left(\frac{z^2}{2i}\right)\right)\overline{\left(5\exp\left(\frac{z^2}{2i}\right)\right)}.
$$
Therefore, a possible choice would be
$$
f(z)=5\exp\left(\frac{z^2}{2i}\right)e^{i\theta},
$$
where $\theta\in\mathbb{R}$ is a fixed parameter that can be figured out by the constraint $f(i)=5$, i.e.,
$$
5=f(i)=5\exp\left(\frac{i}{2}\right)e^{i\theta}\Longrightarrow\theta=-\frac{1}{2}.
$$
Consequently,
$$
f(z)=5\exp\left(\frac{z^2}{2i}\right)e^{-i/2}=5\exp\left(\frac{z^2}{2i}-\frac{i}{2}\right)=5\exp\left(\frac{z^2+1}{2i}\right).
$$
