I have the analytic function $f=u(x,y)+iv(x,y)$ and that $$u=v+e^x\cos(y)+e^{2x}\cos(2y)+\ldots+e^{nx}\cos(ny)$$ where $n\ge2$ and the value $f(0)=\frac{1-i}{2}n$.
After verifying that u and v are harmonic, using the Cauchy-Riemann equations and the initial value to find the constant, I get that $$f(z)=\frac{1-i}{2}\sum_{k=1}^ne^{kz}$$ Now I have to prove that the solutions to the equation $$f(z)\frac{1-e^z}{e^z}=1$$ are purely imaginary. How do I do this?
For $z \ne 0$ we have (finite geometric series !)
$$\sum_{k=1}^ne^{kz}=e^z \frac{1-e^{nz}}{1-e^z}.$$
Can you proceed ?