Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

Put $ z = e^{i \theta} \\ cos\theta = \frac{z^2 + 1}{2-z} \\ d\theta = \frac{dz}{iz} $

The equation can be written as,

$ \int_c \frac{1}{3 + 2(z^2+1/2z)} \frac{dz}{iz} \\ = \int_c \frac{z}{3z+z^2+1} \frac{dz}{iz} \\ = \int_c \frac{dz}{i(z^2+3z+1)} \\ = \frac{1}{i}\int_c \frac{dz}{(z^2+3z+1)} $

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

For …

Put $ z = e^{i \theta} \\ cos\theta = \frac{z^2 + 1}{2-z} \\ d\theta = \frac{dz}{iz} $

The equation can be written as,

$ \int_c \frac{1}{3 + 2(z^2+1/2z)} \frac{dz}{iz} \\ = \int_c \frac{z}{3z+z^2+1} \frac{dz}{iz} \\ = \int_c \frac{dz}{i(z^2+3z+1)} \\ = \frac{1}{i}\int_c \frac{dz}{(z^2+3z+1)} $

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

For …