Obtain Taylor's and laurent series expression for

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

indicating region of convergence.

$\begin{align} f(z)& = \frac{z-1}{(z^2-z-3)}\ f(z)& = \frac{z-1}{(z+1)(z-3)} = \frac{A}{z+1} + \frac{B}{z-3}\ z-1 &= A(z-3)+b(z+1)\ put z&=-1 \ -2 &= -4A\ A&=\frac12\ put z&=3 \ 2& = 4B\ B&=\frac12\ \therefore \frac{z-1}{(z+1)(z_3)} & = \frac{1/2}{(z+1)} + \frac{1/2}{(z-3)} - - - - 1 \ \end{align}$ Hence f(z) is not analytic at z z=-1 & z=3 $\therefore$ f(z) is analytic in 1. $\lvert z \lvert \gt 1$ 2. $1 \gt \lvert z \lvert \gt 3$ 3. $\lvert z \lvert \lt 3$ Case 1. When $\lvert z \lvert \lt 1$ , we get $\lvert z \lvert \lt 3$ by using equation 1 $\begin{align} f(z)&= \frac12(\frac{1}{1+z})+\frac{1}{2(-3)}(\frac{1}{1-\frac z3})\ &= \frac12(1+z)^{-1}-\frac{1}{6}(1-\frac z3)^{-1}\ &= \frac12[1-z+z^2-z^3+- - - ]-\frac{1}{6}[1+\frac z3+\frac {z^2}{9}+ - - -]\ &= \frac13-\frac59 z + \frac{13}{27} z^2\ \end{align}$ This is the required Taylor series case 2. $1 \gt \lvert z \lvert \gt 3$, We get $\lvert \frac1z \lvert \lt 1 \& \lvert \frac z3 \lvert \lt 1$ from equation 1 $\begin{align} f(z)&= \frac12(\frac{1}{1+z})+\frac{1}{2}(\frac{1}{z-3})\ &= \frac{1}{2z}(1+z)^{-1}-\frac{1}{6}(1-\frac z3)^{-1}\ &= \frac{1}{2z}[1-\frac 1z +\frac{1}{z^2}-\frac{1}{z^3}+- - - ]-\frac{1}{6}[1+\frac z3+\frac {z^2}{9}+ - - -]\ &= \frac{1}{2}[1-\frac 1z +\frac{1}{z^2}-\frac{1}{z^3}+- - - ]-\frac{1}{6}[1+\frac z3+\frac {z^2}{9}+ - - -]\ \end{align}$ This is the required laurent series. Case 3. $\lvert z \lvert \gt 3, \ clearly \ \lvert z \lvert \gt 1$ $\frac{ \lvert z \lvert}{3} \gt 1 \& \ \frac{ \lvert z \lvert}{1} \gt 1$ $\frac{3}{ \lvert z \lvert} \lt 1 \& \ \frac{1}{ \lvert z \lvert}{1} \lt 1$ by using equation 1 $\begin{align} f(z)&= \frac{1}{2z} \frac{1}{1+\frac1z} + \frac{1}{2z} \frac{1}{1-\frac3z}\ &= \frac{1}{2z} [1+\frac1z] + \frac{1}{2z} [1-\frac3z]\ &= \frac{1}{2z}[1-\frac 1z +\frac{1}{z^2}-\frac{1}{z^3}+- - - ]+\frac{1}{2z}[1+\frac 3z+\frac{9}{z^2}+\frac{27}{z^3} - - -]\ &= \frac{1}{2z}[2+\frac 2z+\frac {10}{z^2}+\frac{26}{z^3}+ - - -]\ &= [2+\frac 1z+\frac {1}{z^2}+\frac{5}{z^3}+\frac{13}{z^4}+ - - -]\ \end{align}$

This is Required laurent series