Finding an entire $g(z)$ such that $e^{g(z)}=\frac{e^{2z}-1}{\sin(iz)}$.

Question

Take $g(z)=\text{Log}\left(\dfrac{e^{2z}-1}{\sin(iz)}\right)$. Both $e^{2z}-1$ and $\sin(iz)$ have same (simple) zeroes, so $\dfrac{e^{2z}-1}{\sin(iz)}$ is an entire function with no zeroes. Therefore, $\text{Log}\left(\dfrac{e^{2z}-1}{\sin(iz)}\right)$ is entire since the composition of two analytic functions is analytic. But I'm not sure if this is correct since this seems too easy.

Answer 1

I think you are expected to find a simpler expression for $g$.

$\frac {e^{2z}-1} {\sin (iz)}=\frac {-2i (e^{2z}-1)} {e^{z}-e^{-z}}$. If you multiply numerator an denominator by $e^{z}$ you see that this becomes $-2ie^{z}$ so just write $-2i$ as $e^{c}$ to finish. [Answer: $g (z)=z+\ln 2 -i\frac {\pi} 2$].