Let

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

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

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

(z-1)=A(z-3)+B(z+1)

put z= -1

-2 = -4A

A=1/2

Put z=+3

2=4B

B=1/2

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

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

f(z) is analytic in

i)|z| < 1

ii)1<|z|<3

iii)|z|>3

Case 1) when |z|<1 we get |z|<3

by using equation (1)

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

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

=$\frac{1}{2}[1+z]^{-1}-\frac{1}{6}[1-\frac{z}{3}]^{-1}$

=$\frac{1}{2}[1-z+z^2-z^3+....]-\frac{1}{6} [1+\frac{z}{3}+\frac{z^2}{3^2}]$

=$\frac{1}{3}-\frac{5}{9}z+\frac{13}{27}z^2$

This is required taylor series.

case 2) when 1<|z|<3

we get $|\frac{1}{z}|\lt1 \ and \ |\frac{z}{3}| \lt1$

from eqn (1)

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

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

=$\frac{1}{2z}[1+\frac{1}{z}]^{-1}-\frac{1}{6}[1-\frac{z}{3}]^{-1}$

= $\frac {1}{2z}[1-\frac{1}{z}+\frac{1}{z^2}-\frac{1}{z^3}+---]-\frac{1}{6}[1+\frac{z}{3}+\frac{z^2}{9}+\frac{z^3}{27}+----]$

= $\frac {1}{2}[\frac{1}{z}-\frac{1}{z^2}+\frac{1}{z^3}-\frac{1}{z^4}+---]-\frac{1}{6}[1+\frac{z}{3}+\frac{z^2}{9}+\frac{z^3}{27}+----]$

this is required laurent series.

case 3) when |z|>3,clearly |z|>1

$\frac{|z|}{3} \gt 1 \ and \ \frac{|z|}{1} \gt1$

$\therefore \frac{3}{|z|} \lt 1 \ and \ \frac{1}{|z|} \lt1$

By using equation (1)

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

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

= $\frac {1}{2z} [1+\frac {1}{z}]^{-1}+\frac {1}{2z}[1-\frac {3}{z}]^{-1}$

= $\frac {1}{2z} [1-\frac {1}{z}+\frac {1}{z^2}-\frac {1}{z^3}+......]+\frac {1}{2z}[1+\frac {3}{z}+\frac {9}{z^2}+\frac {27}{z^3}+....]$

=$\frac {1}{2z}[2+\frac {2}{z}+\frac {10}{z^2}+\frac{26}{z^3}+....]$

=$\frac {1}{z}+\frac {1}{z^2}+\frac {5}{z^3}+\frac {13}{z^4}$

This is required laurent series...