I can't prove that $f$ admits the real part $u$

Question

Given $v(x,y)=\cfrac{\sinh2y}{\cos2x+\cosh2y}$ and $f\left(\cfrac{\pi}{4}\right)=1$, find the holomorphic function $f(z)=u(x,y)+iv(x,y)$.

I've tried to prove that $v$ is harmonic but I didn't succeed.

Answer 1

With some effort, you can check that $v$ is harmonic. You may use a computer algebra system, but sometimes it doesn't automatically apply the following identities: $$\cos^2(2x)+\sin^2(2x)=1$$ and $$\cosh^2(2y)-\sinh^2(2y)=1.$$ Using the above identities and Cauchy-Riemann equations, it can be solved (or checked) that $$f(z)=u(x,y)+iv(x,y)$$ with $$u(x,y)=\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}$$ satisfies the condition $f(\frac \pi 4)=1$ and is analytic, except when $x=\frac {\pi}2+n\pi,y=0$. I am not sure how to make it holomorphic.