Problem 20 - Advanced Engineering Mathematics

Question

\(\int_{1-i}^{1+\sqrt{3 i}}\left(\frac{1}{z}+\frac{1}{z^{2}}\right) d z, C\) is any contour in the right half-plane \(\operatorname{Re}(z)>0\)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understanding the Integral

We are tasked with evaluating the complex integral \(\int_{1-i}^{1+\sqrt{3}i}\left(\frac{1}{z}+\frac{1}{z^{2}}\right) dz\). This integral is over a curve \(C\) in the right half-plane \(\operatorname{Re}(z)>0\). For complex functions, examining properties such as singularities is crucial in determining how to approach the integral.

Step 2: Check for Singularities

The integrand is \(\frac{1}{z} + \frac{1}{z^2}\). Singularities occur where the function is not defined, specifically at \(z = 0\) for both \(\frac{1}{z}\) and \(\frac{1}{z^2}\). Since the contour lies in the right half-plane and 0 is on the imaginary axis, singularities do not affect the integral, and the integrand is holomorphic within the contour.

Step 3: Application of Cauchy's Integral Theorem

According to Cauchy's Integral Theorem, if a function is holomorphic (analytic) throughout a simply connected domain, then the integral over a closed contour within that domain is zero. Our contour \(C\) is within the right half-plane, where \(\frac{1}{z} + \frac{1}{z^2}\) is holomorphic. Therefore, the integral evaluates to zero along any contour in the region \(\operatorname{Re}(z)>0\).

Step 4: Evaluating the Integral

Based on Cauchy's theorem and the absence of any singularity inside our contour, the integral \(\int_{1-i}^{1+\sqrt{3}i}\left(\frac{1}{z}+\frac{1}{z^{2}}\right) dz\) is zero. As long as the contour \(C\) remains in the right half-plane, the theorem holds, confirming our solution.

Key Concepts

Complex Integration

Complex integration is the process of integrating a complex-valued function over a contour in the complex plane. It’s similar to real integration, but it involves the complexities of complex numbers and functions. In the context of our example, we are dealing with an integral over a curve in the right half-plane, specifically

• from \(1-i\) to \(1+\sqrt{3}i\).
• The integrand \(\frac{1}{z} + \frac{1}{z^2}\) involves rational functions, which are common in complex integration.

Complex integration considers not just the values of the function along the path but also how it behaves around singularities. It requires us to look for points where the function might not be well-defined. This knowledge helps to decide the best method to evaluate the integral, employing the rich framework of complex analysis, including the use of residues and theorems such as Cauchy's Integral Theorem.

Cauchy's Integral Theorem

Cauchy's Integral Theorem is a fundamental result in complex analysis that simplifies the process of evaluating complex integrals. It states that if a function is holomorphic over a simply-connected domain, then the integral of that function over a closed curve within that domain is zero.

In our exercise:

• The function \(\frac{1}{z} + \frac{1}{z^2}\) is holomorphic in the right half-plane.
• Although both components have singularities at \(z = 0\), our contour lies in the region \(\text{Re}(z) > 0\), thus avoiding the singularity.
• This falls precisely into the conditions where Cauchy's Integral Theorem applies, confirming that our integral can be evaluated as zero. The beauty of Cauchy's Theorem is that it allows us to make strong claims about the integrals without directly evaluating them.

Holomorphic Functions

Holomorphic functions, also known as analytic functions, play a significant role in the study of complex analysis. These are functions that are differentiable at every point in their domain, which makes them much nicer and more predictable than merely continuous functions. Differentiability in complex functions involves satisfying the Cauchy-Riemann equations.

In regard to our example:

• The integrand \(\frac{1}{z} + \frac{1}{z^2}\) is holomorphic in the right half-plane, where the contour lies.
• The only potential singular point is at \(z = 0\), which is not included within the contour.
• The fact that the integrand is holomorphic allows us to use powerful results like Cauchy's Integral Theorem to simplify our calculations.

Holomorphic functions are integral to complex integration because their properties lead to elegant solutions and deeper understanding of complex structures.