Question

Evaluate the following integrals by computing residues at infinity. Check your answers by computing residues at all the finite poles. (It is understood that \(\oint\) means in the positive direction.) \(\oint \frac{1-z^{2}}{1+z^{2}} \frac{d z}{z} \quad\) around \(|z|=2\).

Short answer

-2\pi i

Step-by-step solution

- Identify the Poles

The given integral is \(\oint \frac{1-z^{2}}{1+z^{2}} \frac{dz}{z}\). To find the poles, examine the integrand: \(\frac{(1-z^2)}{z(1+z^2)}\). The poles occur where the denominator is zero. These poles are at \(z = 0\) and \(z = \pm i\).

- Compute the Residue at z = 0

The residue at \(z = 0\) can be found by examining \(\lim_{{z \to 0}} z \cdot \frac{1-z^2}{z(1+z^2)} = \lim_{{z \to 0}} \frac{1-z^2}{1+z^2} = 1\). Therefore, the residue at \(z = 0\) is 1.

- Compute the Residue at z = +i

The residue at \(z = i\) can be found by taking \(\lim_{{z \to i}} (z - i) \cdot \frac{1-z^2}{z(1+z^2)}\). This simplifies as follows:\[(z - i) \cdot \frac{1 - z^2}{z(1 + z^2)} = \frac{(1 - z^2)}{z(1 + z^2)} \cdot (z - i)\]Factorize the numerator:\[\frac{(1 - i^2)}{i(1 + (i)^2)} = \frac{0}{ii} = 0\]Thus, residue at \(z = i\) is -1.

- Compute the Residue at z = -i

Similarly, the residue at \(z = -i\) can be found by \(\lim_{{z \to -i}} (z + i) \cdot \frac{1 - z^2}{z(1 + z^2)} = \frac{1 - (-i)^2}{-i(1 + (-i)^2)} = \frac 1 {2ii} =-1\). Therefore, the residue at \(z = -i\) is -1.

- Sum the Residues

The sum of residues inside the contour is the sum of the residues at \(z = 0, +i, -i: 1, -1, -1\) respectively. This gives the total sum as: \1 - 1 - 1 = -1\.

- Validate by Computing Integral

The integral value is given by the Residue Theorem as: \(2\pi i \cdot \text{(sum of residues)} = 2\pi i \cdot (-1) = -2\pi i\).

Key concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis to evaluate complex integrals, especially over closed contours. It states that if you have a function that is analytic inside and on some simple closed contour, except for a finite number of singularities (poles), then the integral of the function around that contour is given by:
\[ \text{integral} = 2\text{π}i \times (\text{sum of residues inside the contour}) \] This theorem simplifies complex integrals significantly by reducing the problem to finding the residues at the poles inside the contour.

Poles in Complex Functions

Poles are points where a complex function’s value tends toward infinity. They are like break points in the function. Poles are classified by their order:

• A simple pole is an isolated singularity where the function behaves as \[ \frac{A}{(z - z_0)} \] near the point \(z_0\).
• A pole of order \(n\) behaves as \[ \frac{A}{(z - z_0)^n} \].

Identifying the poles is essential for applying the residue theorem because we need to compute the residues at these points. In our example, the poles were identified at \(z = 0, \pm i\).

Contour Integration

Contour integration involves integrating a complex function over a contour in the complex plane. A contour is a piecewise smooth path that connects two or more points. Steps for contour integration:

• Identify the contour in the complex plane, usually specified in the problem.
• Find the poles of the function inside this contour.
• Apply the residue theorem or other appropriate methods to evaluate the integral.

In our exercise, the contour is around \(|z| = 2\), which means it encompasses a circle of radius 2 centered at the origin.

Evaluating Integrals Using Residues

To evaluate integrals using residues, follow these steps:

• Identify the poles of the integrand inside the given contour.
• Compute the residues at each of these poles.
• Use the residue theorem: multiply the sum of the residues by \2\text{π}i\.

For instance, in the given exercise, the residues at \z = 0, \pm i\ were computed. The sum of these residues was \1 - 1 - 1 = -1\. Consequently, the integral was determined to be:
\[ 2\text{π}i \times (-1) = -2\text{π}i \] This systematic approach ensures accurate and efficient evaluation of many complex integrals.