Question

b) Using the method of residues, evaluate the following integral:
∫0 to 2π{dθ/](3+2cosθ)}

Answer 1

Answer on Question #85330 – 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

According to the main theorem of residues

$$\oint_{C} f(z) \, dz = 2\pi i \sum_{k=1}^{n} \operatorname{res}_{z = z_k} f(z), \qquad z \in D.$$

where $C$ is the contour of area $D$, $z_k$ are singular points.

Use the substitution $z = e^{i\theta}$ and use the Euler formulas:

$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \qquad i.e. \quad \cos\theta = \frac{1}{2} \left(z + \frac{1}{z}\right);$$ $$3 + 2\cos\theta = 3 + z + \frac{1}{z} = \frac{z^2 + 3z + 1}{z};$$ $$dz = i e^{i\theta} \, d\theta \quad \rightarrow \quad d\theta = \frac{dz}{iz}.$$

Segment $[0; 2\pi]$ changing variables can be thought of as changing $\arg z$ points $z$ belonging to the circle. Indeed, the substitution $z = e^{i\theta}$ translates the segment $[0; 2\pi]$ to the circle $|z| = 1$, $0 \leq \arg z \leq 2\pi$.

We use all the above

$$\int_{0}^{2\pi} \frac{d\theta}{3 + 2\cos\theta} = \oint_{|z|=1} \frac{z}{z^2 + 3z + 1} \frac{dz}{iz} = \frac{1}{i} \oint_{|z|=1} \frac{1}{z^2 + 3z + 1} \, dz$$

The singular points of the integrand are the zeros of the denominator, that is, the roots of the equation $z^2 + 3z + 1 = 0$. These are points $z_1 = \frac{-3 + \sqrt{5}}{2}$ and $z_2 = \frac{-3 - \sqrt{5}}{2}$. Then the denominator can be written as $(z - z_1)(z - z_2)$. Point $z_2$ does not belong to the domain $|z| < 1$ and point $z_1$ belongs and $z_1$ is a pole of the 1st order. Find the residue at the point $z = z_1$, which is a pole of the first order:

$$\begin{aligned} & \operatorname{res}_{z = z_1} \frac{1}{(z - z_1)(z - z_2)} = \lim_{z \to z_1} \frac{1}{(z - z_1)(z - z_2)} (z - z_1) = \lim_{z \to z_1} \frac{1}{(z - z_2)} = \frac{1}{z_1 - z_2} = \frac{1}{\sqrt{5}} \\ & \quad \frac{1}{i} \oint_{|z|=1} \frac{1}{z^2 + 3z + 1} \, dz = \frac{1}{i} \cdot 2\pi i \cdot \operatorname{res}_{z = z_1} \frac{1}{(z - z_1)(z - z_2)} = \frac{2\pi}{\sqrt{5}}. \end{aligned}$$

Answer:

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

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