Problem 17 - Complex Analysis for Mathematics and Engineering

Question

Evaluate \(\int_{0} \cos z d z\), where \(C\) is the polygonal path from 0 to \(1+i\) that consists of the line segments from 0 to 1 and 1 to \(1+i\).

Step-by-Step Solution

Answer

The integral equals \( \sin(1) + i \cos(1) \sinh(1) + \sin(1)(\cosh(1) - 1) \).

Step 1: Understand the Path

The path consists of two segments: from 0 to 1 (along the real axis) and then from 1 to \(1+i\) (moving vertically). The integral will be split into two parts along these segments.

Step 2: Parametrize the First Segment

The first segment is from 0 to 1. Parametrize this path by setting \( z = t \), where \( t \) goes from 0 to 1. Therefore, \( dz = dt \).

Step 3: Integrate Along the First Segment

Substitute the parametrization of the first segment into the integral: \( \int_{0}^{1} \cos(t) \, dt \). This is a standard definite integral with respect to \( t \).

Step 4: Compute the First Integral

Integrate \( \int_{0}^{1} \cos(t) \, dt \). The antiderivative of \( \cos(t) \) is \( \sin(t) \). Evaluate it from 0 to 1: \( \sin(1) - \sin(0) = \sin(1) \).

Step 5: Parametrize the Second Segment

The second segment goes from 1 to \(1+i\). Parametrize this by setting \( z = 1 + i t \), where \( t \) goes from 0 to 1. Therefore, \( dz = i \, dt \).

Step 6: Integrate Along the Second Segment

Substitute the parametrization of the second segment into the integral: \( \int_{0}^{1} \cos(1 + i t) \, i \, dt \). This requires evaluating a complex cosine function along the path.

Step 7: Simplify the Second Integral

Use the fact that \( \cos(1 + i t) = \cos(1) \cosh(t) - i \sin(1) \sinh(t) \) from the identity \( \cos(a + ib) = \cos(a)\cosh(b) - i\sin(a)\sinh(b) \). Substitute this into the integral.

Step 8: Compute the Second Integral

The integral \( \int_{0}^{1} (i[\cos(1)\cosh(t) - i\sin(1)\sinh(t)]) \, dt \) becomes two separate integral parts: \( i\cos(1) \int_{0}^{1} \cosh(t) \, dt \) and \( \sin(1) \int_{0}^{1} \sinh(t) \, dt \). Compute these using the known antiderivatives.

Step 9: Evaluate and Combine Results

The antiderivative of \( \cosh(t) \) is \( \sinh(t) \), evaluated from 0 to 1 to get \( \sinh(1) \). The antiderivative of \( \sinh(t) \) is \( \cosh(t) \), evaluated from 0 to 1 to get \( \cosh(1) - 1 \). Thus the integral is \( i \cos(1) \sinh(1) + \sin(1)(\cosh(1) - 1) \). Add this to the result from the first segment.

Key Concepts

Line Integrals

In complex analysis, we use line integrals to find the accumulation of a function along a curve or path in the complex plane. Simply put, when you calculate a line integral, you're summing up a whole bunch of infinitely small values, each representing the function's value over a tiny piece of the path.
Line integrals have various applications, such as evaluating work done by a force field or finding the circulation of a vector field.

• This is useful in physics and many engineering fields.
• In mathematics, it helps us understand the properties of functions over different paths in the complex plane.

When dealing with complex functions, the key is to break down the path into known segments. For example, in our exercise, the path is divided into two straight-line segments. We then find the integral over each segment separately, taking care to account for the direction and endpoint of each section. This segmented approach makes the calculation manageable and allows us to use suitable parametrizations to evaluate each part.

Complex Integration

Complex integration allows us to extend the concept of integrating real-valued functions to complex-valued ones. When performing complex integration, the idea is to integrate a complex-valued function along a path in the complex plane. This requires careful attention to the real and imaginary components of the function.
While such integrals may initially sound daunting, they're crucial in understanding complex functions' behavior and transformations.

• Completing complex integrals involves breaking the calculation down into real and imaginary parts.
• We may often use complex identities and properties, like Euler's formula or trigonometric identities, to simplify the integrals.
• We will often find ourselves performing real-valued integrals along with calculations to complete the solution.

In the given exercise, handling the complex cosine function along the path involves using trigonometric and hyperbolic identities to re-express the function in simpler forms for integration. Breaking down the real and imaginary parts allows us to evaluate them separately and recombine for the final result.

Parametrization of Paths

Parametrization of paths is critical because it helps us to express a path in terms of a single variable, usually chosen to make calculations easier. In complex problems, understanding how to properly parametrize the path allows you to apply line integrals correctly and efficiently.
This process involves picking a suitable variable, like \( t \), that traces the path from its start to finish.

• Choosing the right parametrization makes integrals easier to evaluate.
• Clear bounds of integration can be set once parametrization is applied.
• This also helps to express complex derivatives in simpler forms.

In our exercise, we parametrize each segment separately: one path from 0 to 1 and another from 1 to \( 1+i \). By applying a variable \( t \) to represent these journeys, we convert each segment into a solvable form. Thus, we translate a potentially complex-looking integral into standard real integrals over known intervals.