Find the Laurent series in $1<|z|<2$ ring

Question

I just started Laurant series, I have this function

$$f(z) = {z^4 + 1 \over {(z-1)(z+2)}}$$ in $1<|z|<2$ ring, I know the theorems,but still cant figure it out, Any help with idea or transformation , welcome!

Answer 1

Since $z^4+1=(z^2-z+3)(z-1)(z+2)-5z+7$, you have\begin{align}\frac{z^4+1}{(z-1)(z+2)}&=z^2-z+3+\frac{-5z+7}{(z-1)(z+2)}\\&=z^2-z+3+\frac2{3(z-1)}-\frac{17}{3(z+2)}.\end{align}On the other hand, since $1<|z|<2$, you have$$\frac1{z-1}=-\frac1{1-z}=z^{-1}+z^{-2}+z^{-3}+\cdots$$and$$\frac1{z+2}=\frac12\cdot\frac1{1+\frac z2}=\frac12-\frac z{2^2}+\frac{z^2}{2^3}-\frac{z^3}{2^4}+\cdots$$Now, put it all together.

Answer 2

\begin{align} f(z) & =z^2-z+3+\frac{2}{3(z-1)}-\frac{17}{3(z+2)} \\ & = z^2-z+3-\frac 23 \cdot \frac{1}{1-z}-\frac{17}{6} \cdot \frac{1}{1-(-z/2)} \\ & = z^2-z+3-\frac 23 \big(1+z+z^2+z^3 \cdots \big)-\frac{17}{6}\big(1-\frac 12z+\frac 14z^2-\frac18 z^3+\cdots \big) \\ & = -\frac 12-\frac 14z-\frac 38z^2+\cdots \end{align}