Find Laurent’s series which represent the function

$f(z) = \frac{2}{(z-1)(z-2)}$ when $(i) |z| \lt 1 (ii) 1 \lt |z| \lt 2 (iii) |z| \gt 2$

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

Case (i) |z|<1 ∴|z|<2

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

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

Case(ii) 1< |z|<2

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

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

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

Case(iii) |z|>2 ∴|z|>1 , we write

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

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

=$ \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 +….] $