Find all maximal rings centered at $0$ where $\displaystyle f(z)=\frac{5-2i}{z^2-(5+2i)z+10i}$ has a Laurent series and compute them.
What I've got so far:
Since $f$ is the quotient of two polynomials it is holomorphic on rings not containing the zeros of its denominator. We have $\displaystyle f=\frac{1}{z-5}-\frac{1}{z-2i}$ so the rings are $K_{0,2}(0),\ K_{2,5}(0)$ and $K_{5,\infty}(0)$. But now I don't know how to go on. Looking at other threads it seems helpful to use the geometrical series. Using this I can obtain
$\displaystyle f=\frac{5-2i}{z-2i}\cdot\frac{1}{z-5}=\frac{5-2i}{z-2i}\cdot\frac{1}{1+(z-6)}=\frac{5-2i}{z-2i}\cdot\sum_{n=0}^{\infty}(-1)^n(z-6)^n$ for $|z-6|<1$
I don't see how this leads me anywhere. How do I obtain the laurent series on said rings?
For $|z|<2$, we can write
$$-\left(\frac{1}{z-i2}\right)=\frac1{ i2}\left( \frac{1}{1-(z/i2)}\right)=\frac1{ i2}\sum_{n=0}^\infty \left(\frac{z}{i2}\right)^n\tag1$$
while for $|z|>2$, we can write
$$-\left(\frac{1}{z-i2}\right)=-\frac1{ z}\left( \frac{1}{1-(i2/z)}\right)=-\frac1z\sum_{n=0}^\infty \left(\frac{i2}{z}\right)^n\tag2$$
For $|z|<5$, we can write
$$\left(\frac{1}{z-5}\right)=-\frac15\left( \frac{1}{1-(z/5)}\right)=-\frac15\sum_{n=0}^\infty \left(\frac{z}{5}\right)^n\tag3$$
while for $|z|>5$, we can write
$$\left(\frac{1}{z-5}\right)=\frac1z\left( \frac{1}{1-(5/z)}\right)=\frac1z\sum_{n=0}^\infty \left(\frac{5}{z}\right)^n\tag4$$
For $|z|<2$, use $(1)$ and $(3)$.
For $2<|z|<5$, use $(2)$ and $(3)$.
For $|z|>5$, use $(2)$ and $(4)$.