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

$f(z)=\frac{3z-3}{2z^2-5z+2}=\frac{3(z-1)}{2(z-5/4)^2-9/8}=3/2 \frac{z-1}{(z-5/4)^2-9/16}$

$=-3/2 \frac{z-1}{9/16-(z-5/4)^2}=-3/2(\frac{z-5/4+1/4}{9/16-(z-5/4)^2}) = -3/2(\frac{z-5/4}{9/16-(z-5/4)^2}+\frac{1/4}{9/16-(z-5/4)^2})$

$=-3/2(\frac{z-5/4+1/4}{9/16-(z-5/4)^2}) = -3/2(\frac{16/9(z-5/4)}{1-16/9(z-5/4)^2}+2/9\frac{1}{1-16/9(z-5/4)^2})$

Now try to think of this as follows:

$\frac{1}{1-x}=1+x+x^2+...x^n$

$\frac{1}{1+x}=1-x+x^2-...$

$\frac{1}{1-x^2}=1/2 ( \frac{1}{1-x}+\frac{1}{1+x} )=1+x^2+...+x^{2n}$

Also:

$\frac{x}{1-x^2}=\frac{1+x-1}{1-x^2}=\frac{1+x}{1-x^2}-\frac{1}{1-x^2}=\frac{1}{1-x}-\frac{1}{1-x^2}$

And now apply above calculus and series expansion...

$x=\sqrt{\frac{16}{9}}(z-5/4)=\frac{4}{3}(z-\frac{5}{4})$

Above equations now transform to:

$= -3/2(\frac{\frac{4}{3}x}{1-x^2}+2/9\frac{1}{1-x^2})=$

$= -3/2(\frac{4}{3}\frac{x}{1-x^2}+2/9\frac{1}{1-x^2})=$

And now apply upper equations:

$= -\frac{3}{2}(\frac{4}{3}(\frac{1}{1-x}-\frac{1}{1-x^2})+\frac{2}{9}\frac{1}{1-x^2})=$

$= -\frac{3}{2}( (-\frac{4}{3}+\frac{2}{9})\frac{1}{1-x^2}+\frac{4}{3}\frac{1}{1-x})$

$=\frac{3}{2}(\frac{4}{3}-\frac{2}{9})\frac{1}{1-x^2}-2\frac{1}{1-x}$

$=\frac{3}{2}(\frac{4}{3}-\frac{2}{9}) \frac{1}{1-\frac{16}{9}(z-5/4)^2}-2\frac{1}{1-\frac{4}{3}(z-5/4)}$

$=\frac{5}{3}\frac{1}{1-(\frac{4}{3}(z-5/4))^2}-2\frac{1}{1-\frac{4}{3}(z-5/4)}$

Answer 2

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 $