Question: Obtain Laurent series for $f(z) = \frac{4z-3}{z^2-z-6}$ at z = 1
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 209 views
 modified 7 months ago  • written 11 months ago by
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Centre = (1,0)

$\frac{4z+3}{z^2 -z-6} = \frac{4z-3}{z^2 - 3z + 2z -6} = \frac{4z-3}{z(z-3) + 2(z-3)} = \frac{4z-3}{(z+2)(z-3)}$

$\frac{4z-3}{(z+2)(z-3)} = \frac{A}{z+2} + \frac{B}{z-3} \\ 4z-3 = A(z-3) + B(z+2) \\ 12 - 3 = B (5) \hspace{0.50cm} [z = 3] \\ B = \frac{9}{5}$

Put z = -2, $\therefore A = \frac{11}{5}$

$\frac{4z-3}{(z+2)(z-3)} = \frac{11}{5}(\frac{1}{z+2}) + \frac{9}{5}(\frac{1}{z-3}) \\ = \frac{11}{5}[\frac{1}{(z-1) + 3}] + \frac{9}{5}[\frac{1}{(z-1) - 2}]$

case(i): |z-1| $\lt$ 2

$\frac{|z-1|}{2} \lt 1\\ \frac{|z-1|}{3} \lt 1 \\ f(z) = \frac{11}{5}[\frac{1}{(z-1) + 3}] + \frac{9}{5}[\frac{1}{(z-1) - 2}] \\ = \frac{11}{5 \times 3}[\frac{1}{(z-1)/3 + 1}] + \frac{9}{5 \times 2}[\frac{1}{(z-1)/2 - 1}] \\ = \frac{11}{15}[1 - (\frac{z-1}{3}) + (\frac{z-1}{3})^2 - (\frac{z-1}{3})^3 + ...] - \frac{9}{10}[1 + (\frac{z-1}{2}) + (\frac{z-1}{2})^2 + (\frac{z-1}{2})^3 + ...]$

case(ii): 2 $\lt$ |z-1| $\lt$ 3

$\frac{2}{|z-1|} \lt 1 \\ \frac{|z-1|}{3} \lt 1 \\ f(z) = \frac{11}{5}[\frac{1}{(z-1) + 3}] + \frac{9}{5}[\frac{1}{(z-1) - 2}] \\ = \frac{11}{5 \times 3}[\frac{1}{(z-1)/3 + 1}] + \frac{9}{5(z-1)}[\frac{1}{1 - 2/(z-1)}] \\ = \frac{11}{15}[1 - (\frac{z-1}{3}) + (\frac{z-1}{3})^2 - (\frac{z-1}{3})^3 + ...] + \frac{9}{5(z-1)}[1 + \frac{2}{z-1} - (\frac{2}{z-1})^2 + (\frac{2}{z-1})^3 + ....]$

case(iii): |z-1| $\gt$ 3

$3 \lt |z-1| \\ \frac{3}{|z-1|} \lt 1 \\ \frac{2}{|z-1|} \lt \frac{3}{|z-1|} \lt 1 \\ \therefore \frac{2}{|z-1|} \lt 1 \\ f(z) = \frac{11}{5}[\frac{1}{(z-1) + 3}] + \frac{9}{5}[\frac{1}{(z-1) - 2}] \\ = \frac{11}{5(z-1)}[\frac{1}{1 + 3/(z-1)}] + \frac{9}{5(z-1)}[\frac{1}{1 - 2/(z-1)}] \\ = \frac{11}{5}[\frac{1}{(z-1) + 3}][1 - (\frac{3}{z-1}) + (\frac{3}{z-1})^2 - (\frac{3}{z-1})^3 + ....] + \frac{9}{5(z-1)}[1 + (\frac{2}{z-1}) + (\frac{2}{z-1})^2 + (\frac{2}{z-1})^3 + ....]$