Problem 2 - Advanced Engineering Mathematics

Question

(a) Choosing the line \(y=\frac{1}{3} x, 0 \leq x \leq 3\) we have \(z=x+\frac{1}{3} x i, d z=\left(1+\frac{1}{3} i\right) d x .\) Thus \\[ \int_{C} e^{z} d z=\int_{0}^{3} e^{\left(1+\frac{1}{3} i\right) x}\left(1+\frac{1}{3} i\right) d x=\left.e^{\left(1+\frac{1}{3} i\right) x}\right|_{0} ^{3}=e^{3+i}-e^{0}=\left(e^{3} \cos 1-1\right)+i e^{3} \sin 1 \\] (b) \(\int_{C} e^{z} d z=\int_{0}^{3+i} e^{z} d z=\left.e^{z}\right|_{0} ^{3+i}=e^{3+i}-e^{0}=\left(e^{3} \cos 1-1\right)+i e^{3} \sin 1\)

Step-by-Step Solution

Answer

Both paths yield the same result: \((e^3 \cos 1 - 1) + ie^3 \sin 1\).

Step 1: Understanding the given path

We start by identifying the path for the integral. The line given is \( y = \frac{1}{3}x \), with \( 0 \leq x \leq 3 \). This path is represented in the complex plane as \( z = x + \frac{1}{3}xi \).

Step 2: Finding the differential dz

By differentiating \( z = x + \frac{1}{3}xi \) with respect to \( x \), we have \( dz = \left(1 + \frac{1}{3}i\right)dx \). This replaces \( dz \) in the integral.

Step 3: Substituting into the integral

Now substitute \( dz = \left(1 + \frac{1}{3}i\right)dx \) into the integral: \[ \int_{0}^{3} e^{\left(1+\frac{1}{3}i\right)x}\left(1+\frac{1}{3}i\right) dx \]

Step 4: Solving the integral

Integrate \( e^{\left(1+\frac{1}{3}i\right)x} \) with respect to \( x \). The indefinite integral is \( \left. e^{\left(1+\frac{1}{3}i\right)x} \right|_0^3 \). Evaluate it from \( 0 \) to \( 3 \), which gives:\[ e^{3+i} - e^0 = e^{3+i} - 1 \]

Step 5: Expressing in rectangular form

Express \( e^{3+i} \) in rectangular form using Euler's formula: \[e^{3+i} = e^3(\cos 1 + i\sin 1) \]. Therefore, the result is:\[ (e^3 \cos 1 - 1) + ie^3 \sin 1 \]

Step 6: Verifying with alternate path

Similarly, if calculated with the direct path \( C \) as a straight line in the complex plane from \( 0 \) to \( 3 + i \), the integral remains:\[ \int_0^{3+i} e^z dz = e^{3+i} - e^0 = (e^3 \cos 1 - 1) + ie^3 \sin 1 \]

Key Concepts

Complex Exponential Functions

Complex exponential functions form the foundation of combining exponential growth with rotational motion. At its core, the complex exponential function is an extension of the familiar real exponential function into the complex plane. We define it as:

• For any complex number \( z = x + yi \), the complex exponential is given by: \[ e^z = e^{x+yi} = e^x e^{yi} \]
• The real part \(x\) governs the magnitude, while the imaginary part \(yi\) introduces rotation.

Due to its ability to represent continuous growth and circular motion, complex exponentials are crucial in studying oscillatory systems such as waves and alternating current circuits. In our given problem, we explore the path defined by an exponential function along a complex plane line defined by \( y = \frac{1}{3}x \).

Euler's Formula

Euler's formula is a pivotal bridge connecting complex exponentials to trigonometric functions. It is expressed as:\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]

• This formula provides a powerful way to express complex numbers, allowing rotations in the complex plane.
• For example, if \( z = 3 + i \) and using Euler's formula, we have: \[ e^{3+i} = e^3(\cos 1 + i\sin 1) \]

Euler's formula is inherently useful in simplifying complex integrals, like in our exercise, transforming the expression of \( e^{3+i} \) into a combination of real and imaginary components. This conversion is especially helpful to visualize and interpret the impact of both magnitude and angle within the context of complex functions.

Rectangular Form of Complex Numbers

The rectangular form is the most straightforward form for expressing complex numbers, using real and imaginary parts. A complex number \( z \) is often represented as: \[ z = a + bi \] where \(a\) is the real part and \(b\) is the imaginary part.

• The rectangular form provides a direct way to plot complex numbers on a Cartesian plane, with \(a\) corresponding to the x-coordinate and \(b\) to the y-coordinate.
• In our problem, once we solve the integral, the result \( (e^3 \cos 1 - 1) + ie^3 \sin 1 \) is expressed in this format.

By converting the exponential result to its rectangular form using Euler’s formula, we enhance our understanding of the integral’s outcome and its geometrical interpretation on the complex plane, illustrating changes in both magnitude and direction.