# Complex functions and poles

Suppose we have a complex function $$f(z)$$ with simple poles on the complex plane, and we know exactly where these poles are located (but we don't know how the function depends on z) Is there any way to build up the exact form of $$f(z)$$ just from its poles?

Svein
Suppose we have a complex function $$f(z)$$ with simple poles on the complex plane, and we know exactly where these poles are located (but we don't know how the function depends on z) Is there any way to build up the exact form of $$f(z)$$ just from its poles?
No. You can conclude that the function is of the form $H(x)+\sum_{n}\frac{c_{n}}{(z-z_{n})^{k_{n}}}$, where H is analytic, zn are the poles and the cn and kn are constants (the kn are positive integers).

mfb
Mentor
For every function f(z), 2*f(z), z*f(z) and so on have the same poles (the last one assuming 0 is not a pole).

Yes, thanks for both replies, I think the first one is related with Mittag Leffers theorem, the poles I am talking about follow a law, for example, a harmonic series on the real axis. But the sum over all this poles gives a complicated function the complex plane. To be clear, I am trying to evaluate a function of the type: $$f(z)=\sum_{n=-\infty}^{+\infty}\frac{1}{z-z_{n}}$$ where the zn poles can follow a law, for example $$z_{n}=A*(n+\frac{1}{2})^{2}$$ where A is constant. The problem is that the series depending on these poles can give as a result really complicated functions of z, I just wanted to know if there is an analytical way to carry out these sums.

mfb
Mentor
Some poles will match (e. g. n=1, n=-2).

A sum like$$\sum_{n=1}^\infty \frac{1}{1+(n+c)^2}$$ should have an analytic result, and you can modify your sum to have that shape.

Thats actually what I have done, and yes it has to have an analytical form, I was just wondering about an analytical method to arrive at it, rather than using Mathematica, just to know if there exist a general method or something, many thanks for the replies

Svein
Picked up my trusty Ahlfors (Complex analysis) and he shows that $\sum_{n=-\infty}^{\infty}\frac{1}{(z-n)^{2}}=\frac{\pi^{2}}{\sin^{2}\pi z}$ and $\frac{1}{z}+\sum_{n\neq 0}(\frac{1}{z-n}+\frac{1}{n})=\pi\cot(\pi z)$. Might give you some ideas.
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