Answer to Question #294432 in Calculus for Lucky

Question

Find all possible Taylor's series and Laurent series expansions of f(z) = (2z-3) /(z-2) (z-1) about z=0?

Answer 1

Solution:

Consider $f(z)=\frac{z-1}{z^{2}-2 z-3}=\frac{z-1}{(z-3)(z+1)}$

Using partial fraction we get

 $\frac{z-1}{(z-3)(z+1)}=\frac{a}{z-3}+\frac{b}{z+1}$

i.e. z-1=a(z+1)+b(z-3)

put z=3 , we get $a=\frac{1}{2}$

put z=-1 , we get $b=\frac{1}{2}$

$\frac{z-1}{(z-3)(z+1)}=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1}$

Hence f(z) is not analytic at z=-1 and z=3

$\therefore$ f(z) is analytic in

$* |z|<1 \\ * 1<|z|<3 \\ * |z|>3$

(i)

$|z|<1 \\\because|z|<1,|z|<3$

$\begin{aligned} &\therefore f(z)=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1} \\ &=\frac{1}{2} \cdot \frac{1}{-3} \cdot \frac{1}{1-\frac{z}{3}}+\frac{1}{2} \frac{1}{1+z} \\ &=-\frac{1}{6}\left(1-\frac{z}{3}\right)^{-1}+\frac{1}{2}(1+z)^{-1} \end{aligned}$

$\begin{aligned} &=-\frac{1}{6}\left(1+\frac{z}{3}+\left(\frac{z}{3}\right)^{2}+\left(\frac{z}{3}\right)^{3}+\ldots \ldots\right)+\frac{1}{2}\left(1-z+z^{2}-z^{3}+\ldots \ldots\right) \\ &f(z)=\frac{1}{3}-\frac{5}{9} z+\frac{13}{27} z^{2}-\ldots \ldots \end{aligned}$

Above is required Taylor's series

(ii) 1<|z|<3

We get $\left|\frac{1}{z}\right|<1$  and $\left|\frac{z}{3}\right|<1$

$\begin{aligned} &\therefore f(z)=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1} \\ &=\frac{1}{2} \cdot \frac{1}{-3} \cdot \frac{1}{1-\frac{z}{3}}+\frac{1}{2} \cdot \frac{1}{z} \cdot \frac{1}{1+\frac{1}{z}} \\ &=\frac{1}{-6} \cdot \frac{1}{1-\frac{z}{3}}+\frac{1}{2 z} \cdot \frac{1}{1+\frac{1}{z}} \\ &=-\frac{1}{6}\left(1-\frac{z}{3}\right)^{-1}+\frac{1}{2 z}\left(1+\frac{1}{z}\right)^{-1} \\ &=-\frac{1}{6}\left(1+\frac{z}{3}+\left(\frac{z}{3}\right)^{2}+\left(\frac{z}{3}\right)^{3}+\ldots \ldots\right)+\frac{1}{2 z}\left(1-\left(\frac{1}{z}\right)+\left(\frac{1}{z}\right)^{2}-\left(\frac{1}{z}\right)^{3}+\ldots\right)^{2}+\frac{1}{2}\left(\frac{1}{z}-\left(\frac{1}{z}\right)^{2}+\left(\frac{1}{z}\right)^{3}-\left(\frac{1}{z}\right)^{4}+\ldots\right. \\ &f(z)=-\frac{1}{6}\left(1+\frac{z}{3}+\left(\frac{z}{3}\right)^{2}+\left(\frac{z}{3}\right)^{3}+\ldots\right) \end{aligned}$

This is required Laurent's series

(iii) |z|>3

$\begin{aligned} &\because|z|>3, \text { clearly, }|z|>1 \\ &\therefore\left|\frac{z}{3}\right|>1 \text { and }|z|>1 \\ &\text { i.e. }\left|\frac{3}{z}\right|<1 \text { and }\left|\frac{1}{z}\right|<1 \\ &\therefore f(z)=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1} \\ &=\frac{1}{2 z} \cdot \frac{1}{1-\frac{3}{z}}+\frac{1}{2 z} \cdot \frac{1}{1+\frac{1}{z}} \\ &=\frac{1}{2 z}\left(1-\frac{3}{z}\right)^{-1}+\frac{1}{2 z}\left(1+\frac{1}{z}\right)^{-1} \\ &=\frac{1}{2 z}\left(1+\left(\frac{3}{z}\right)+\left(\frac{3}{z}\right)^{2}+\left(\frac{3}{z}\right)^{3}+\ldots\right)+\frac{1}{2 z}\left(1-\left(\frac{1}{z}\right)+\left(\frac{1}{z}\right)^{2}-\left(\frac{1}{z}\right)^{3}+\ldots .\right) \\ &=\frac{1}{z}+\frac{1}{z^{2}}+\frac{5}{z^{3}}+\frac{13}{z^{4}}+\ldots \cdots \cdots \\ &f(z)=\frac{1}{z}+\frac{1}{z^{2}}+\frac{5}{z^{3}}+\frac{13}{z^{4}}+\ldots \ldots \ldots . \end{aligned}$

This is the required Laurent's series