Problem 15 - Complex Analysis for Mathematics and Engineering

Question

Show that \(\int_{C} 1 d z=z_{2}-z_{1}\), where \(C\) is the line segment from \(z_{1}\) to \(z_{2}\), by parametrizing \(C .\)

Step-by-Step Solution

Answer

The integral \(\int_{C} 1 \, dz = z_2 - z_1\).

Step 1: Understand the Problem

We are given a line integral along the path \(C\), which is the line segment from \(z_1\) to \(z_2\). We need to find the integral of the function 1 with respect to \(z\) and show that it equals \(z_2 - z_1\) by using parametrization of \(C\).

Step 2: Parametrize the Line Segment

The line segment from \(z_1\) to \(z_2\) can be parametrized as \(z(t) = (1-t)z_1 + tz_2\), where \(t\) varies from 0 to 1. This gives us a way to express any point on the line segment in terms of \(t\).

Step 3: Differentiate the Parametrization

Differentiate the parametrization with respect to \(t\) to find \(\frac{dz}{dt}\). We have \(\frac{dz}{dt} = z_2 - z_1\).

Step 4: Set Up the Line Integral

The line integral \(\int_C 1 \, dz\) over the path from \(t = 0\) to \(t = 1\) becomes: \[\int_{0}^{1} 1 \, \frac{dz}{dt} \, dt.\]

Step 5: Compute the Line Integral

Substitute \(\frac{dz}{dt} = z_2 - z_1\) into the integral: \[\int_{0}^{1} (z_2 - z_1) \, dt = (z_2 - z_1) \int_{0}^{1} 1 \, dt.\] This simplifies to \((z_2 - z_1)\).

Step 6: Evaluate the Integral

Evaluate the integral \(\int_{0}^{1} 1 \, dt = 1\). Therefore, the line integral is \((z_2 - z_1) \cdot 1 = z_2 - z_1\).

Key Concepts

Line Integral

A line integral, in its simplest form, is an integral where the function is evaluated along a curve instead of over an interval on the real line. It is particularly useful in complex analysis for evaluating functions over paths in the complex plane.
Let's break it down:

• Instead of integrating over an axis, like you would in typical calculus problems, we integrate along a path or curve.
• The curve can be anything from a straight line to a complex winding path.
• The function being integrated can also be a complex-valued function of a complex variable.

Line integrals have practical applications, including calculating work done by a force along a path in physics.

Parametrization

Parametrization is the process of expressing a curve through a parameter, usually denoted as \( t \), which varies over an interval. In our context, we are parametrizing the line segment from \( z_1 \) to \( z_2 \).
Here's how it works practically:

• We use a parameter, \( t \), which ranges from 0 to 1.
• The line segment can be represented as \( z(t) = (1-t)z_1 + tz_2 \).
• This formula allows us to find any point on the line segment by plugging in a value for \( t \).

Parametrization is essential because it transforms our curve into a form that is easier to manage when performing integration and differentiation.

Integration

Integration, specifically in this context, involves finding the integral of a function over the path defined by our parametrization.
Let's dissect this process:

• We parameterized the curve, so now we focus on integrating over the parameter \( t \).
• The original line integral \( \int_C 1 \, dz \) reformulates to an integral over \( t \): \( \int_{0}^{1} 1 \, \frac{dz}{dt} \, dt \).
• Integration simplifies \( \int_0^1 1 \, dt \) to 1 since we are integrating a constant over \( t \).

Ultimately, this means if you multiply by \( dz/dt = z_2 - z_1 \), the result of the integration is simply \( z_2 - z_1 \), which confirms what we set out to prove.