# Function poles and divergence of series

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.

• The series expansion is invalid for $|f|\ge 1$ and the pole is at $f=1$. Commented May 20, 2015 at 15:47
• I tought that since they are equal elsewhere, the series divergence point is the pole of the function, so they have the same residue there. It seems is not that easy.
– user115760
Commented May 20, 2015 at 16:09

The function $g(z) = 1/(1-f(z))$ is holomorphic when $f(z) \ne 1$, but the series converges only when $|f(z)| < 1$. In particular, because of the Maximum Modulus Theorem, if you have a closed contour $\gamma$ on which $|f(z)| < 1$, you will also have $|f(z)| < 1$ inside $\gamma$, and no singularities of $g$ there. Thus you can never use the series to find a residue of $g$.
However, what is true is that if $p$ is a point where $f(p) = 1$, then $f(z) = 1 + f'(p) (z - p) + O((z-p))^2$ as $z \to p$, so $$g(z) = \dfrac{1}{-f'(p)(z-p) + O((z-p)^2)} = -\dfrac{1}{f'(p) (z-p)} + O(1)$$ and thus the residue of $g$ at $p$ is $-1/f'(p)$.