Problem 27 - Advanced Engineering Mathematics

Question

\(\int_{C}\left(z^{2}+4\right) d z\), where \(C\) is the line segment from \(z=0\) to \(z=1+i\)

Step-by-Step Solution

Answer

The integral is \(4 + \frac{16}{3}i\).

Step 1: Parameterize the Line Segment

The first step in solving this problem is to parameterize the path \( C \) from \( z = 0 \) to \( z = 1 + i \). Since this is a straight-line segment, we can use the parameter \( t \) such that \( z(t) = t(1 + i) \) where \( t \) ranges from \( 0 \) to \( 1 \). This implies that for \( t = 0 \), \( z(0) = 0 \), and for \( t = 1 \), \( z(1) = 1 + i \).

Step 2: Compute the Derivative \( dz \)

To compute the differential \( dz \) from our parameterization, take the derivative of \( z(t) = t(1 + i) \) with respect to \( t \). Thus, \( \frac{dz}{dt} = 1+i \), and consequently, \( dz = (1+i) \, dt \).

Step 3: Substitute into the Integral

Substitute the parameterization into the integral and replace \( dz \) with \( (1+i) dt \). The integral becomes: \[ \int_{0}^{1} \left( (t(1+i))^2 + 4 \right) (1+i) \, dt. \] Simplifying further, we compute \( (t(1+i))^2 = t^2(1+2i-i^2) = t^2(2i+2) \), thus the integral becomes: \[ (1+i) \int_{0}^{1} ( (2+2i) t^2 +4 ) dt. \]

Step 4: Evaluate the Integral

Now, let's evaluate the integral \( \int_{0}^{1} ((2+2i)t^2 + 4) dt \). Compute it term by term: 1. \( \int_{0}^{1} (2+2i) t^2 \, dt = \frac{(2+2i)}{3} \).2. \( \int_{0}^{1} 4 \, dt = 4 \).Substitute back to get the integral: \[ (1+i) \left( \frac{(2+2i)}{3} + 4 \right) = (1+i) \left( \frac{(2+2i) + 12}{3} \right) = (1+i) \cdot \frac{2 + 2i + 12}{3} = (1+i) \cdot \frac{14 + 2i}{3}. \]

Step 5: Simplify the Result

Finally, distribute \( (1+i) \) across the expression: \[ (1+i) \cdot \frac{14 + 2i}{3} = \frac{1}{3} (14 \cdot (1+i) + 2i \cdot (1+i)) = \frac{1}{3} (14 + 14i + 2i - 2) \]. This simplifies to: \[ \frac{1}{3} \cdot (12 + 16i) = \frac{12}{3} + \frac{16i}{3} = 4 + \frac{16}{3}i. \]

Key Concepts

Parameterization

The journey through complex integration begins with understanding parameterization. When dealing with line integrals along a path, it's necessary to describe the path in terms of a parameter.
In this problem, the path is a straight line from the origin to the point \( z = 1 + i \). A straight line segment can be neatly parameterized by introducing a parameter \( t \), where it ranges from 0 to 1.
Our path is then

• \( z(t) = t(1 + i) \)
• Here, \( t = 0 \) gives \( z(0) = 0 \)
• \( t = 1 \) gives \( z(1) = 1 + i \)

So, this parameterization effectively transforms our problem into something simpler to manage. It's like simplifying a route on a map to a straight line trajectory.

Line Integrals

Line integrals are a powerful toolkit in complex analysis. They help evaluate a function over a complex curve or path. Unlike regular integrals that sum values over an interval in the real plane, line integrals account for values across a curvy slice of the complex plane.
In our exercise, the integral \( \int_{C} (z^2 + 4) \, dz \) embodies this principle. We're summing contributions of the function \( z^2 + 4 \) over the parameterized line path.

• For line integrals, we parameterize the line segment \( C \).
• The derivative \( dz \) is essential, found by differentiating parameter \( t \).
• Substitute this parameterization and \( dz \) into the function being integrated.

This simplifies tackling the integral across complex paths, rendering them tangible in evaluations.

Complex Analysis

Complex analysis opens up a richer mathematical landscape by treating complex numbers as more than just sums of real and imaginary parts.
With complex integration, a major element of complex analysis, we can explore how functions vary over complex paths. The key here is recognizing how derivatives, parameterization, and substituting into integrals leverage the multi-dimensional nature of complex numbers.

• Complex functions can be integrated across curves or lines, as in the problem.
• Substituting parameterized paths allows real calculations in the complex plane.
• Complex derivatives and their influence reshape integral least squares evaluations.

Complex analysis provides a powerful lens to view calculus beyond real numbers.

Integral Evaluation

Evaluating the integral in complex analysis involves breaking down and solving around our parameter.
After parameterizing the path, the next step is computing \( dz \) and substituting this and the parameterized \( z \) into the integral. The integral

• takes on a form simpler to solve.
• Different components inside the integral are tackled separately.
• Results are combined for the final solution.

In our step-by-step solution, each contributing term of the integral was evaluated separately before simplifying to the final answer.
This systematic breakdown allows tackling complex integration challenges with confidence, unveiling results from tangled initial expressions to simplified, useful outcomes.