<s>[INST] Laurent Series of $f(z) = \frac{1}{(1-z)(1-cz)}$ in the point $z_0 = \frac{1}{c}$, where $c = \frac{3}{4}i, z \in \mathrm{C}$. I am currently trying to find the Laurent Series to $f(z) = \frac{1}{(1-z)(1-cz)}$ in the point $z_0 = \frac{1}{c}$. Therefore I calculated the partial fractions and found that $f(z) = \frac{A}{1-z} + \frac{B}{1-cz}$, with $A = \frac{16}{25} + \frac{12}{25}i$ and $B=\frac{9}{25} - \frac{12}{25}i$. Now when I use the geometric sum to find the Laurent Series I get $f(z) = \sum_{k=0}^{\infty} z^k A + \sum_{k=0}^{\infty} (cz)^k B = \sum_{k=0}^{\infty} z^k(A+c^kB)$. But that means that $|cz|$ must be less then 1 and than $|z| < |1/c| = |z_0|$. Doesn't than mean, I can't use that approach or is it not possible to develop the series in that point?
 [/INST] You computed the Laurent series of $f$ in $z_0 = 0$, which then obviously only converges for $|z| < |1/c|$. If you want to compute the Laurent series in $z_0$ = 1/c, you can factorize $f$ as
$$f(z) = \frac{1}{z - \frac{1}{c}} \frac{1}{c(z-1)}.$$
The first factor is in the correct form. Now expand the second factor in $z_0 = \frac{1}{c}$ and you get the desired Laurent series.
EDIT: Because there seems to be some confusion, here is the Laurent series around $z_0 = 1/c$. We have for $0<|z-1/c|<|1-1/c|=5/3$
$$\frac{1}{1-z} = \frac{1}{\frac{c-1}{c} - (z - \frac{1}{c})} = \frac{c}{c-1}\frac{1}{1 - \frac{c}{c-1}(z - \frac{1}{c})} = \frac{c}{c-1}\sum\limits_{n = 0}^{\infty} \frac{c^n}{(c-1)^n}(z-\frac{1}{c})^n.$$
Thus
$$f(z) = (z-\frac{1}{c})^{-1}\frac{1}{1-c}\sum\limits_{n = 0}^{\infty} \frac{c^n}{(c-1)^n}(z-\frac{1}{c})^n = \frac{1}{1-c}\sum\limits_{n = 0}^{\infty} \frac{c^n}{(c-1)^n}(z-\frac{1}{c})^{n-1}.$$
Shifting the index by one, you get
$$f(z) = -\sum\limits_{n = -1}^{\infty} \frac{c^{n+1}}{(c-1)^{n+2}}(z-\frac{1}{c})^n.$$
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