Find Laurent Series Of $\frac{z-7}{z^2+z-2}$ at $1<|z|<2$

Question

$$\frac{z-7}{z^2+z-2}$$ at $1<|z|<2$

$$\frac{z-7}{z^2+z-2}=-\frac{2}{z-1}+\frac{3}{z+2}$$

in $1<|z|<2$ both functions are analytic so it is just taylor series?

Answer 1

No, it's not just the Taylor series.

In the case of $\frac3{z+2}$, it is, indeed. In that case, whe have\begin{align}\frac3{z+2}&=\frac3{2-(-z)}\\&=\frac32\times\frac1{1-\frac{-z}2}\\&=\frac32\left(1-\frac z2+\frac{z^2}{2^2}-\frac{z^3}{2^3}+\cdots\right)\\&=\frac32-\frac3{2^2}z+\frac3{2^3}z^2-\frac3{2^4}z^3+\cdots,\end{align}since $\left|-\frac{-z}2\right|<1$. On the other hand,\begin{align}-\frac2{z-1}&=-\frac2z\times\frac1{1-\frac1z}\\&=-\frac2z\left(1+\frac1z+\frac1{z^2}+\frac1{z^3}+\cdots\right)\\&=-\frac2z-\frac2{z^2}-\frac2{z^3}-\frac2{z^4}\cdots,\end{align}since $|z|>1$.