Question

Using the method of residues, evaluate the following integral:
2π
∫ dθ/3+2cos(θ)
0

Answer 1

Answer on Question #84492 – Math – Complex Analysis

Question

Using the method of residues, evaluate the following integral:

$$\int_{0}^{2\pi} \frac{d\theta}{3 + 2 \cos(\theta)}$$

Solution

Let $z = e^{i\theta}$, $dz = i e^{i\theta} d\theta$, $d\theta = \frac{dz}{iz}$, $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z + \frac{1}{z}}{2}$

The complex $z$ describes the unit circle $C_1$ in the positive sense as $\theta$ varies from 0 to $2\pi$. So, the integral becomes

$$\begin{array}{l} \int_{0}^{2\pi} \frac{d\theta}{3 + 2 \cos(\theta)} = \oint \frac{\frac{dz}{iz}}{3 + 2 \left( \frac{z + \frac{1}{z}}{2} \right)} _{C_1(0)} = -i \oint \frac{dx}{3z + z^2 + 1} _{C_1(0)} = \\ = -i(2\pi i) \sum_{j} \operatorname{Rez}\left(z^2 + 3z + 1, z_j\right) = 2\pi \sum_{j} \operatorname{Rez}\left(z^2 + 3z + 1, z_j\right), \end{array}$$

where the sum of the residues extends over all the poles of $\frac{1}{z^2 + 3z + 1}$ inside the unit disk.

$$z^2 + 3z + 1 = 0$$

The roots are

$$\begin{array}{l} z = \frac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2} = \frac{-3 \pm \sqrt{5}}{2} \\ z_1 = \frac{-3 - \sqrt{5}}{2}, z_2 = \frac{-3 + \sqrt{5}}{2}. \end{array}$$

Only $z_2 = \frac{-3 + \sqrt{5}}{2}$ is inside $C_1(0)$.

Use the formula

$$\operatorname{Rez}_{z = z_0} \frac{g(z)}{h(z)} = \frac{g(z_0)}{h'(z_0)},$$

where $z_0$ is a simple zero of $h(z)$

$$\operatorname{Rez}\left(z^2 + 3z + 1, z_2\right) = \frac{1}{2 \left( \frac{-3 + \sqrt{5}}{2} \right) + 3} = \frac{1}{\sqrt{5}}$$

Hence

$$\int_{0}^{2\pi} \frac{d\theta}{3 + 2 \cos(\theta)} = \frac{2\pi}{\sqrt{5}}.$$

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