Given the real part of an analytic function $f:\mathbb{C}\rightarrow \mathbb{C}$
$$\text{Re}[f] = \frac{\sin(2x)}{\cosh(2y)-\cos(2x)},$$
I have to find the imaginary part of $f$.
I tried using the Cauchy-Riemann equations, but $\frac{\partial\text{Re}[f]}{\partial x}=\frac{\partial\text{Im}[f]}{\partial y}$ gives me the following integral:
$$\text{Im}[f]=2\int\frac{\cos(2x)\cosh(2y)-1}{\left(\cosh(2y)-\cos(2x)\right)^2}dy$$
and here is where the difficulties start. I put it in Wolfram Alpha and yeah, I don't think this is the way to go.
Is there another way to calculate Im$[f]$ without having such crazy integrals in your way?
Using addition formulas you have:
$\cosh(2y)-\cos(2x)=\cos(2iy)-\cos(2x)=2\sin(x+iy)\sin(x-iy)=2\sin(z)\sin(\bar z)$
Applying same kind of transformation to a numerator that would be symmetric, gives:
$\sin(2x)-i\sinh(2y)=\sin(2x)-\sin(2iy)=2\cos(x+iy)\sin(x-iy)=2\cos(z)\sin(\bar z)$
Thus $f(z)=\dfrac 1{\tan(z)}$ is a candidate and since it is analytic, it is the one searched for (have a look at the post below).
Is it Possible to have Two Distinct Analytic Functions with the Same Real Part?