Problem 5 - Complex Analysis for Mathematics and Engineering

Question

Find \(\int_{c}(2 z-1)\left(z^{2}-z\right)^{-1} d z\) for the (a) circle \(C=C_{2}^{+}(0)=\\{z:|z|=2\\}\) having positive orientation. (b) circle \(C=C_{j}^{+}(0)=\left\\{z:|z|=\frac{1}{2}\right\\}\) having positive orientation.

Step-by-Step Solution

Answer

(a) 0 (b) \(-2\pi i\)

Step 1: Analyze the Integral

We need to evaluate the integral \(\int_{C}(2z-1)(z^{2}-z)^{-1} dz\) over given circles. Notice that the integrand is \(\frac{2z-1}{z(z-1)}\). The singularities are at \(z = 0\) and \(z = 1\). These are the poles of the function.

Step 2: Determine Singularities Location

For part (a), we analyze the circle \(|z| = 2\), which contains both singularities \(z = 0\) and \(z = 1\) inside. For part (b), we analyze the circle \(|z| = \frac{1}{2}\), which contains only the singularity \(z = 0\) since \(|1| > \frac{1}{2}\).

Step 3: Apply Residue Theorem for Circle (a)

By the Residue Theorem, for a function with isolated singularities, the integral of \(f(z)\) around a closed curve is \(2\pi i\) times the sum of the residues inside the curve. For \(|z| = 2\), both 0 and 1 are inside, so we find residues at both.

Step 4: Calculate Residue at z=0

The residue at \(z = 0\) is found by multiplying \((2z -1)\) by \(z\), so \(\lim_{{z \to 0}} \frac{2z-1}{z-1} = -1\).

Step 5: Calculate Residue at z=1

Similarly, the residue at \(z = 1\) is determined by \((2z-1)\), so \(\lim_{{z \to 1}} \frac{2z-1}{z} = 1\).

Step 6: Compute Integral for Circle (a)

The sum of residues for \(|z| = 2\) is \(-1 + 1 = 0\). Hence, \(\int_{C}(2z-1)(z^{2}-z)^{-1} dz = 2\pi i \times 0 = 0\).

Step 7: Apply Residue Theorem for Circle (b)

For the circle \(|z| = \frac{1}{2}\), only \(z = 0\) is inside. Thus, we only consider this residue.

Step 8: Compute Integral for Circle (b)

The residue at \(z = 0\) is \(-1\), so the integral is \(2\pi i \times (-1) = -2\pi i\).

Key Concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis that helps evaluate complex integrals of functions with isolated singularities. Specifically, it states that the integral of a function around a closed contour is equal to \(2\pi i\) times the sum of the residues of the function inside that contour.

Here's a quick way to understand it:

• Find the singular points of your function, also known as poles, that lie inside the contour of integration.
• Determine the residues at these poles.
• Sum up these residues.
• Multiply the total residue by \(2\pi i\) to get the value of the integral.

This theorem is especially useful because it converts the sometimes complex problem of contour integration into a simpler algebraic operation of adding residues. Use the Residue Theorem when dealing with functions that are analytic except at isolated singularities.

Contour Integration

Contour integration involves integrating a complex function over a path, or contour, in the complex plane. When you work with contour integration:

• Select the contour over which the integration is to be performed. Common choices include circles, rectangles, or other simple curves.
• Ensure the function is analytic (differentiable) everywhere except possibly at a set of isolated points within or on the contour.
• Apply theorems from complex analysis, like the Residue Theorem, to evaluate the integral along the specified path.

Contour integration is valuable in electromagnetism and fluid dynamics, where it simplifies the calculation of integrals that appear in physical problems. It is also frequently used in solving problems in signal processing and probability theory due to its adaptability in handling complex exponential functions.

Singularities

In complex analysis, singularities refer to points at which a function ceases to be analytic. These can be classified mainly into three types:

• Removable Singularities: A point where a function is undefined but can be redefined such that the function becomes analytic.
• Poles: Points where a function goes to infinity. The order of the pole describes how the function behaves near that point (for example, a simple pole at \(z = a\) is of order one).
• Essential Singularities: Points at which the function's behavior is chaotic and can't be simplified to either of the previous types.

Identifying and analyzing singularities help us use the Residue Theorem effectively. Knowing whether a point is a pole or an essential singularity allows us to calculate residues accurately, as in the original exercise where the poles at \(z = 0\) and \(z = 1\) were crucial for solving the integral.