Problem 15 - Advanced Engineering Mathematics

Question

By partial fractions, \(\oint_{C} \frac{2 z+1}{z(z+1)} d z=\oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z\) (a) By Theorem 18.4 and (4) of Section 18.2 $$\oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z=2 \pi i+0=2 \pi i$$ (b) \(\quad\) By writing \(\oint_{C}=\oint_{C_{1}}+\oint_{C_{2}}\) where \(C_{1}\) and \(C_{2}\) are the circles \(|z|=1 / 2\) and \(|z+1|=1 / 2,\) respectively we have by Theorem 18.4 and (4) of Section 18.2 $$\begin{aligned} \oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z &=\oint_{C_{1}} \frac{1}{z} d z+\oint_{C_{1}} \frac{1}{z+1} d z+\oint_{C_{2}} \frac{1}{z} d z+\oint_{C_{2}} \frac{1}{z+1} d z \\ &=2 \pi i+0+0+2 \pi i=4 \pi i \end{aligned}$$(c) \(\operatorname{since} f(z)=\frac{2 z+1}{z(z+1)}\) is analytic within and on \(C\) it follows from Theorem 18.4 that $$\oint_{C} \frac{2 z+1}{z^{2}+z} d z=0$$

Step-by-Step Solution

Answer

The integral evaluates to \( 2\pi i \).

Step 1: Understand the Problem

We need to evaluate the integral \( \oint_{C} \frac{2z+1}{z(z+1)} \, dz \) using the method of partial fractions. We break down the integrand into simpler parts that can be integrated separately.

Step 2: Break Down into Partial Fractions

Express the integrand \( \frac{2z+1}{z(z+1)} \) as the sum of partial fractions: \( \frac{A}{z} + \frac{B}{z+1} = \frac{2z+1}{z(z+1)} \). Solve for constants \( A \) and \( B \) by equating coefficients.

Step 3: Solve for Constants

Equating coefficients, we get \( Az + A + Bz = 2z + 1 \). Comparing coefficients, \( A + B = 2 \) and \( A = 1 \). Solving, \( B = 1 \). Thus, \( \frac{2z+1}{z(z+1)} = \frac{1}{z} + \frac{1}{z+1} \).

Step 4: Apply Cauchy's Integral Theorem

Using Cauchy's Integral Theorem, evaluate the integrals \( \oint_{C} \frac{1}{z} \, dz \) and \( \oint_{C} \frac{1}{z+1} \, dz \), given that \( z = 0 \) is inside \( C \) and \( z = -1 \) is outside \( C \). The result is \( 2\pi i \) from \( \oint_{C} \frac{1}{z} \, dz \) and zero from \( \oint_{C} \frac{1}{z+1} \, dz \).

Step 5: Sum Individual Integrals

Sum the evaluated integrals to find the total integral: \( \oint_{C} \frac{2z+1}{z(z+1)} \, dz = 2\pi i + 0 = 2\pi i \). This confirms the solution stated in part (a) of the problem statement.

Step 6: Consider the Given Paths

In part (b), note \( C_1 \) and \( C_2 \) are circles of radius \( \frac{1}{2} \). Using these paths and Cauchy's theorem, each of \( \oint_{C_1} \frac{1}{z} \, dz \) and \( \oint_{C_2} \frac{1}{z+1} \, dz \) evaluate to \( 2\pi i \) while the others evaluate to zero as they don't encompass their respective poles.

Step 7: Verify Analyticity

In part (c), \( f(z) = \frac{2z+1}{z(z+1)} \) is analytic on and within \( C \), so by Cauchy’s Integral Theorem, \( \oint_{C} \frac{2z+1}{z^2+z} \, dz = 0 \) because there are no poles within \( C \).

Key Concepts

Partial Fractions

Partial fractions are a powerful tool in calculus and complex analysis, especially useful for simplifying integrands. The main idea is to decompose a complicated rational expression into simpler fractions. In this exercise, we are given the function \( \frac{2z+1}{z(z+1)} \). This expression can be rewritten as the sum of simpler parts: \( \frac{1}{z} \) and \( \frac{1}{z+1} \).
By finding constants that satisfy \( \frac{A}{z} + \frac{B}{z+1} = \frac{2z+1}{z(z+1)} \), we can individually integrate these simpler fractions. Solving gives \( A = 1 \) and \( B = 1 \), allowing a straightforward integration.
This simplification is crucial as it turns complex integrals into manageable calculations, particularly in contour integration.

Cauchy's Integral Theorem

Cauchy's Integral Theorem is a cornerstone of complex analysis. It states that if a function is analytic and its derivative is continuous on some region, then the integral of the function over any closed contour in that region is zero. In simpler terms, when a function has no poles (points of singularity) within a closed curve, the integral along that curve equals zero.
In our exercise, even though the function \( \frac{2z+1}{z(z+1)} \) seems complex, the partial fractions \( \frac{1}{z} \) and \( \frac{1}{z+1} \) allow the use of Cauchy’s theorem. Since these components are analytic in the regions of integration, the integrals can be evaluated directly. Particularly, \( \oint_{C} \frac{1}{z} \, dz \) yields \( 2\pi i \) if the curve \( C \) surrounds the pole at \( z = 0 \), while \( \oint_{C} \frac{1}{z+1} \, dz \) results in zero when \( z = -1 \) is not within the contour.

Analytic Functions

Analytic functions are those that are differentiable at every point in a given region. This characteristic is vital in complex integration since it allows the use of theorems like Cauchy’s. In our exercise, the function \( f(z) = \frac{2z+1}{z(z+1)} \) is expressed in partial fractions as \( \frac{1}{z} + \frac{1}{z+1} \).
Both components are analytic in their defined regions, except at their respective poles. This highlights that outside small exceptions (like these poles), the behavior of an analytic function is predictable and smooth across the region of interest. The ability to express it in terms of partial fractions simplifies understanding and ensures seamless application of integration theorems.

Contour Integration

Contour integration involves integrating complex functions over predefined paths or contours in the complex plane. It's a method widely used in evaluating complex integrals. In this exercise, contours \( C_1 \) and \( C_2 \) are circles around specific points, and help evaluate the integral of \( \frac{2z+1}{z(z+1)} \) as \( \oint_{C} = \oint_{C_1} + \oint_{C_2} \).
When dealing with contour integration, it is crucial to understand the location of poles relative to the contour.

• If a pole is inside the contour, Cauchy's Theorem and Residue Theorem provide straightforward methods for integration, often resulting in non-zero values like \( 2\pi i \).
• If a pole lies outside the contour, the integral evaluates to zero.
• The correct choice of contour ensures efficient problem-solving and highlights the nature of functions within these contours.

This integration method is essential because it permits the evaluation of seemingly complex integrals by knowing just a few properties of the function and its poles.