Help finding the roots of a complex function $f(z)=e^{\cos(z)}-1$ where $z \in \mathbb C$
Solution 1:
You can express the solutions in terms of the (real) inverse hyperbolic sine since \begin{align*} \cos z = 2\pi iq &\Leftrightarrow - i\sinh \left( {i\left( {z + \left( {2p + \tfrac{1}{2}} \right)\pi} \right)} \right) = 2\pi iq \\ & \Leftrightarrow i\left(z+\left( {2p + \tfrac{1}{2}} \right)\pi\right)= - \sinh ^{ - 1} (2\pi q) \\ & \Leftrightarrow z = - \left( {2p + \tfrac{1}{2}} \right)\pi + i\sinh ^{ - 1} (2\pi q). \end{align*} Here $p,q\in \mathbb{Z}$ are arbitrary.