Find the domain $D$ of the function
$ f(z) = \sum (-1)^n(z-1)^n - \sum i^{n-1} (z-i)^n $
Then determine $f(z)$, for all $z \in D$
I figured for the first part of the function i could use the sum of an infinite geometric series and we get:
$ \sum (-1)^n (z-1)^n = \frac{1}{2-z}$ since we can ignore the alternating aspect, as it doesnt interfere with the convergence.
However, I'm stuck on how I'd apply the same method to the second part of the function, and definining the domain.
Any help would be hugely appreciated, thanks!
Note that each sum separately has a radius of convergence of $1$; thus, the domain of convergence is the intersection of the disks $|z-1| \le 1$ and $|z-i| \le 1$.
Within this domain, the sums converge to
$$f(z) = \frac1{1+z-1} + i \frac1{1-i(z-i)} = \frac1z - \frac1{z+i 2} = \frac{i 2}{z (z+i 2)}$$