Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

$ z^2 - 2z - 3 = z^2 - 3z + z - 3 \\ = z(z-3) + 1(z-0) = (z+1)(z-3) $

Now,

$ \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 = 3

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

Put z = -1

$ -1-1 = A(-1-3) \implies A = \frac{1}{2} $

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

There are three regions of convergence

(i) |z| $ \lt $ 1

(ii) 1 $ \lt $ |z| $ \lt $ 3

(iii) |z| $ \gt $ 3

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

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

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

$ 1 \lt |z| \hspace{0.20cm} \& \hspace{0.20cm} |z| \lt 3 \\ \frac{1}{|z|} \lt 1 \hspace{0.20cm} \& \hspace{0.20cm} \frac{|z|}{3} \lt 1 \\ f(z) = \frac{1}{2}(\frac{1}{z+1}) + \frac{1}{2}(\frac{1}{z-3}) \\ = \frac{1}{2(z)} (\frac{1}{1+1/z}) + \frac{1}{2(3)}(\frac{1}{z/3-1}) \\ = \frac{1}{2(z)} (\frac{1}{1+1/z}) - \frac{1}{6}(\frac{1}{1- z/3}) \\ = \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 + ....] $

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

$ 3 \lt |z| \implies \frac{3}{|z|} \lt 1 \implies \frac{1}{|z|} \lt \frac{3}{|z|} \lt 1 \\ \therefore \frac{1}{|z|} \lt 1 \hspace{0.20cm} \& \hspace{0.20cm} \frac{3}{|z|} \lt 1 $

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