Question

Evaluate the integral \(\int_{C} z^{m}\left(z^{*}\right)^{n} d z\), where \(m\) and \(n\) are integers and \(C\) is the circle \(|z|=1\) taken counterclockwise.

Short answer

The integral \(\int_{C} z^{m}\left(z^{*}\right)^{n} d z\) equals \(2\pi i\) if \(m = n+1\) and 0 otherwise.

Step-by-step solution

Replace the complex conjugate

Since the circle is \(|z|=1\), we can replace \(z^{*}\) on the unit circle with \(\frac{1}{z}\), so the integral becomes \(\int_{C} z^{m}\left(\frac{1}{z}\right)^{n} d z\)

Simplify the expression

Simplifying, we have \(\int_{C} z^{m-n} dz\). This expression can be evaluated directly using the formula for a power of \(z\), specifically \(\int_{C} z^{p} dz = 2\pi i\) if \(p = -1\), and 0 otherwise.

Apply formula and evaluate the integral

We have \(p = m - n\). If \(p = -1\) (which means \(m = n+1\)), we get that the integral is \(2\pi i\), and if \(p \neq -1\) (which means \(m \neq n+1\)), the integral is 0.

Key concepts

Contour Integration

Contour integration is a method used in complex analysis to evaluate integrals over paths or curves in the complex plane. Imagine tracing a path along a curve, where each point on this path is represented by a complex number. The integral sums up the values on this path.

Key points to understand:

• The path is called a "contour" and is often chosen to be simple, like circles or lines.
• For our integral, the path is the unit circle, a specific contour where \(|z|=1\).

Contour integration allows us to simplify problems that might be tricky or impossible to solve with real-valued integrals alone.

Unit Circle

The unit circle is an important concept in both algebra and complex analysis. It is defined as the set of all points in the complex plane that have a distance of 1 from the origin.

When working with the unit circle:

• Any complex number \(z\) on this circle satisfies \(|z| = 1\).
• This property simplifies problems in complex analysis, because taking the complex conjugate \(z^*\) for a number on the unit circle just means finding the multiplicative inverse, i.e., \(\frac{1}{z}\).

Utilizing the unit circle in integrals often leads to elegant solutions.

Complex Conjugate

A complex conjugate involves flipping the sign of the imaginary part of a complex number. If you have a complex number \(z = a + bi\), its conjugate is \(z^* = a - bi\).

Interesting facts about complex conjugates:

• They play a crucial role in simplifying integrals over circles, especially the unit circle.
• On the unit circle, as noted earlier, \(z^*\) simplifies to \(\frac{1}{z}\), because when \(|z|=1\), multiplying \(z\) and \(z^*\) results in 1.

This insight is key to solving complex integrals with ease.

Integral Evaluation

Evaluating integrals in complex analysis involves calculating the sum along a path in the complex plane. For our specific integral \(\int_{C} z^{m-n} dz\), the approach depends on the exponent.

Here's how it generally works:

• If the power \(p\) of \(z\) equals \(-1\), meaning \(m - n = -1\), the integral value is \(2\pi i\).
• For any other power, the integral is zero due to the properties of analytic functions and closed curves.

This method simplifies the evaluation considerably, leveraging powerful theorems in complex analysis.