[Math] Laurent series expansion of $f(z)=\frac{3z-3}{2z^2-5z+2}$ in the annulus $\frac{1}{2}\lt|z-1|\lt1$

Question

Find the Laurent series for:

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

in the annulus $\frac{1}{2}\lt|z-1|\lt1$

I know it requires a bit of fiddling then using the formula for geometric expansion, but I'm struggling to get my head round this one. Thanks!

Answer 1

Firstly, ${3z-3}\over {2z^2-5z=2}$=$1\over {2z-1}$+$1\over {z-2}$ Because ${1\over {2|z-1|}}<1$,hence we have$${1\over {2z-1}}={1\over {2(z-1)+1}}={1\over{{2(z-1)}(1+{1\over {2(z-1)}}})} ={1\over {2(z-1)}}(\sum_{0}^{\infty}({-1\over{2(z-1)}})^{n})$$ on the other hand, since |z-1|<1,then we will have$${1\over {(z-2)}}={-1\over {1-(z-1)}}=-\sum_{0}^{\infty}(z-1)^n $$ thus,we have the final result:$\sum_{0}^{\infty}$$(-1)^n\over {2^{n+1}(z-1)^{n+1}}$$-\sum_{0}^{\infty}(z-1)^n $