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Obtain Tayiors & Laurents series expansions of function $f(z) = \frac{z-1}{z^2-2z-3}$ indicating regions of convergence.
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Let $f(z) = \frac{z-1}{(z+1)(z-3)} = \frac{(1/2)}{(z+1)}+\frac{(1/2)}{(z-3)}$

Possible regions of convergence about z=0 are (i) |z|<1 (ii) 1< |z|<3 (iii) |z|>3

Case (i) |z|<1 ∴|z|<3

$f(z) = \frac{1}{2}(1+z)^{-1} - \frac{1}{6} (1-\frac{z}{3})^{-1}$

$f(z)= \frac{1}{2} [ 1- z + z^2 -z^3 +z^4 …..] – \frac{1}{6} [1+ \frac{z}{3} +(\frac{z}{3})^2 + (\frac{z}{3})^3….]$

Case(ii) 1< |z|<3

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

$f(z)= \frac{1}{2z}(1+\frac{1}{z})^{-1} \,– \frac{1}{6}(1-\frac{z}{3})^{-1}$

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

Case(iii) |z|>3 ∴|z|>1 , we write

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

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

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