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Question: Obtain Laurent series for $ f(z) = \frac{2z - 3}{z^2 - 4z + 3} $ in powers of z-4 indicating regions of convergence in each case.
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 150 views
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modified 4 months ago  • written 7 months ago by gravatar for Manan Bothra Manan Bothra0
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$ \frac{2z-3}{z^2 -4z+3} = \frac{2z-3}{(z-1)(z-3)} $

Consider

$ \frac{2z-3}{(z-1)(z-3)} = \frac{A}{z-1} + \frac{B}{z-3} \\ 2z - 3 = A(z-3) + B(z-1) \\ 2(1) - 3 = A(1-3) \hspace{0.50cm} [z = 1] \\ A = \frac{1}{2} $

Put z = 3, $\therefore B = \frac{3}{2}$

$ \frac{2z-3}{(z-1)(z-3)} = \frac{1}{2}(\frac{1}{z-1}) + \frac{3}{2}(\frac{1}{z-3}) \\ = \frac{1}{2}(\frac{1}{(z-4)+3}) + \frac{3}{2}(\frac{1}{(z-4) +1}) \\ $

case(i): |z-4| $\lt$ 1

$ \frac{|z-4|}{3} \lt \frac{1}{3} \lt 1 \\ \therefore \frac{|z-4|}{3} \lt 1\\ f(z) = \frac{1}{2}(\frac{1}{3((z-4)/3+1)}) + \frac{3}{2}(\frac{1}{(z-4) +1}) \\ = \frac{1}{6}[1 - \frac{z-4}{3} + (\frac{z-4}{3})^2 - (\frac{z-4}{3})^3 + ....] + \frac{3}{2}[1 - (z-4) + (z-4)^2 - (z-4)^3 + ....] $

case(ii): 1 $\lt$ |z-4| $\lt$ 3

$ \frac{1}{|z-4|} \lt 1 \\ \frac{|z-4|}{3} \lt 1 i.e. \frac{1}{|z-4|} \lt 1 \\ \frac{|z-4|}{3} \lt 1 \\ f(z) = \frac{1}{2}[\frac{1}{3((z-4)/3 + 1)}] + \frac{3}{2(z-4)}[\frac{1}{(1/(z-4) + 1)}] \\ = \frac{1}{6}[1 - \frac{z-4}{3} + (\frac{z-4}{3})^2 - (\frac{z-4}{3})^3 + ....] + \frac{3}{2(z-4)}[1 - \frac{1}{z-4} + (\frac{1}{z-4})^2 - (\frac{1}{z-4})^3 + ...] $

case(iii): |z-4| $\gt$ 3

$ 3 \lt |z-4| \\ \frac{3}{|z-4|} \lt 1 \implies \frac{1}{|z-4|} \lt 1 \\ \therefore \frac{1}{|z-4|} \lt 1 \\ \frac{3}{|z-4|} \lt 1 \\ f(z) = \frac{1}{2(z-4)}[\frac{1}{1 + 3/(z-4)}] + \frac{3}{2(z-4)}[\frac{1}{1 + 1/(z-4)}] \\ = \frac{1}{2(z-4)} [1 - \frac{3}{z-4} + (\frac{3}{z-4})^2 - (\frac{3}{z-4})^3 + ...] + \frac{3}{2(z-4)} [1 - \frac{1}{z-4} + (\frac{1}{z-4})^2 - (\frac{1}{z-4})^3 + ....] $

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modified 4 months ago  • written 4 months ago by gravatar for Manan Bothra Manan Bothra0
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