given is a function g, which assigns to every z of complex numbers the value 1/((z-1)*(z-i)) I have to show that it is possible to denote the value above as a limit of a geometric sequence $\sum_{k=0}^\infty = \frac{1}{(z-i)*(z-1)}$. For example, the previous question was z to 1/(1-z), which could be described as the usual limit of an ordinary geometric sequence. What strucks me here is the imaginary unit, as I do not know how to eliminate it from the sequence. Do you have any tips? I tried to do partial fractions but did not manage to continue from there.
I will put my solution on here and hope someone can correct any mistakes:
Starting with $\frac{1}{(z-1)(z-i)}$ I use partial fractions and get $\frac{1}{(1-i)}\left(\frac{1}{(z-1)}-\frac{1}{(z-i)}\right)$ I rewrite that as $\frac{i}{(1-i)}\left(\frac{-1}{(1-z)}(\frac{1}{i})-\frac{1}{1+iz})\right)$ I then write the first expression as $\left(\frac{-1}{1-i}\right)\left(\frac{1}{1-z}\right)=\left(\frac{-1}{1-i}\right)\sum_{n=0}^\infty z^n$ minus the second, which is term $\left(\frac{i}{1-i}\right)\sum_{n=0}^\infty (-iz)^n$. So combined it would be $$\left(\frac{-1}{1-i}\right)\sum_{n=0}^\infty z^n - \left(\frac{i}{1-i}\right)\sum_{n=0}^\infty (-iz)^n$$ Am I right?
Kind Regards