Yesterday I tried to calculate the residues of a function the way below, but soon I realized it won't work. Now I have a question about the poles of a function, and a series representing it. $$z\in \Bbb C\setminus \{a\in \Bbb C\,|\,f(a)=1\} \Rightarrow 0\le |f(z)|\lt1\ $$ Edit: the criteria above is not possible according to the anwser. $$g(z)={1\over 1-f(z)}=\sum_{k=0}^\infty f(z)^k$$ $g(z)$ diverges only when $f(z)=1$, $f(z)$ is holomorphic on the whole complex plane. So $g(z)$ is an infinite sum of function powers, and it has singularities at isolated points $a$ where $f(a)=1$.
However, since $f(z)$ is holomorphic everywhere: $$\oint _\gamma f(z)^k \,dz= 0$$ $$\oint_\gamma\left(\sum_{k=0}^\infty f(z)^k\right)\,dz=\sum_{k=0}^\infty \oint _\gamma f(z)^k \,dz=0=\oint_\gamma g(z)\,dz$$
It is not always true according to the residue theorem.
If I have a function just like $g(x)$, given as an infinite sum of holomorphic functions, and I know where it diverges, but cannot get a closed form, is it possible to calculate the residues of the poles, or even can I say that a series have poles? (I am mainly interested in function like the $f(x)$ and $g(x)$ above)
ps: I am still a high school student, only learned complex analysis by myself so probably I missed out some parts of the subject, but I cannot find anything about it.