# Complex function to Taylor and Laurent series

I am trying to express a function with Taylor and Laurent series. I've been reading my textbook and also various online resources, but I still can't follow any of the example problems. Here's what I understand so far.

I have a function of a complex-valued number z, and its denominator is $$0$$ at $$z= 1$$ and $$2$$.

$$f(z)=\frac{5-z}{z^2-3z+2}$$

From this, I think I should have three separate series. One for $$|z|<1$$, one for $$1<|z|<2$$, and one for $$|z|>2$$. I think I managed to get the first one by deriving the Taylor series for $$f$$, but I know that this only has a radius of convergence of $$1$$.

How can I proceed to derive the other two series? One resource I read stated that when $$z$$ is greater than the radius of convergence, I can use the fact that $$1/z to work with a geometric series, but I am confused on the algebraic manipulation.

Any feedback/constructive criticism is appreciated.

Note that$$\frac{5-z}{z^2-3z+2}=\frac3{z-2}-\frac4{z-1}$$and so, if $$1<\lvert z\rvert<2$$, then\begin{align}f(z)&=-\frac3{2-z}+\frac4{1-z}\\&=-\frac{\frac32}{1-\frac z2}+\frac4{1-z}\\&=-\frac32\sum_{n=0}^\infty\left(\frac z2\right)^n-4\sum_{n=-\infty}^{-1}z^n\text{ (since 1<\lvert z\rvert<2)}\\&=-3\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}-4\sum_{n=-\infty}^{-1}z^n.\end{align}Can you deal with the case $$\lvert z\rvert>2$$ now?
• Here, I am just using the fact that$$\frac1{1-z}=\begin{cases}\sum_{n=0}^\infty z^n&\text{ if }\lvert z\rvert<1\\-\sum_{n=-\infty}^{-1}z^n&\text{ if }\lvert z\rvert>1.\end{cases}$$ – José Carlos Santos Feb 18 '19 at 18:34
You'll have a partial fraction expansion in terms of $$1/(z-1)$$ and $$1/(z-2)$$ so you'll need Laurent expansions for each. Use that $$\frac1{z-a}=-\frac1 a\frac1{(1-z/a)}=\sum_{n=0}^\infty-\frac{z^n}{a^{n+1}}$$ if $$|z|<|a|$$ and $$\frac1{z-a}=\frac1 z\frac1{(1-a/z)}=\sum_{n=1}^\infty a^nz^{-n-1}$$ if $$|z|>|a|$$.