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 $1/z<R$ to work with a geometric series, but I am confused on the algebraic manipulation.
Any feedback/constructive criticism is appreciated.
Best Answer
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?