Problem 12 - Advanced Engineering Mathematics

Question

\(\int_{1-i}^{1+2 i} z e^{z^{2}} d z\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} e^{-3 + 4i} - \frac{1}{2} e^{1 - 2i} \).

Step 1: Understand the Integral

We need to evaluate the contour integral \( \int_{1-i}^{1+2i} z e^{z^{2}} d z \). The integral is over a path from the complex number \(1-i\) to \(1+2i\). This suggests that a useful method is evaluating the integral using the Fundamental Theorem of Calculus for contour integrals.

Step 2: Find an Antiderivative

Notice that the derivative of \( F(z) = \frac{1}{2} e^{z^2} \) is \( F'(z) = z e^{z^2} \) by using the chain rule. Thus, \( F(z) = \frac{1}{2} e^{z^2} \) serves as an antiderivative of our integrand \( z e^{z^2} \).

Step 3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus for contour integrals tells us that if \( F(z) \) is an antiderivative of \( f(z) \), then \( \int_{a}^{b} f(z) \, dz = F(b) - F(a) \). For our problem, we have:\[\int_{1-i}^{1+2i} z e^{z^2} \, dz = \left[ \frac{1}{2} e^{z^2} \right]_{1-i}^{1+2i}.\]

Step 4: Evaluate the Antiderivative at the Limits

Calculate \( F(1+2i) \) and \( F(1-i) \):- \( F(1+2i) = \frac{1}{2} e^{(1+2i)^2} = \frac{1}{2} e^{(1 - 4 + 4i)} = \frac{1}{2} e^{-3 + 4i} \)- \( F(1-i) = \frac{1}{2} e^{(1-i)^2} = \frac{1}{2} e^{(1 - 2i)} = \frac{1}{2} e^{1-2i} \)

Step 5: Evaluate the Integral

Now substitute back:\[\int_{1-i}^{1+2i} z e^{z^2} \, dz = \frac{1}{2} e^{-3 + 4i} - \frac{1}{2} e^{1 - 2i}.\]Using Euler's formula \( e^{a+bi} = e^a(\cos(b) + i\sin(b)) \), both terms can be expressed in exponential form, but a numerical approximation or simplification can also be considered the final result.

Key Concepts

Complex Analysis

Complex analysis is an area of mathematics that deals with complex numbers and functions of a complex variable. This discipline extends the concepts of calculus to complex numbers. Complex numbers are composed of a real part and an imaginary part, and they are expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit (\( i^2 = -1 \)).

• **Functions of Complex Variables:** In complex analysis, functions can be defined for complex numbers, and these functions can have derivatives and integrals, just like functions of real numbers.
• **Contour Integrals:** A contour integral is a type of complex integral where you integrate a function along a path in the complex plane, referred to as a contour or path.

In the original exercise, \( \int_{1-i}^{1+2i} z e^{z^{2}} d z \) represents a contour integral. The path from \( 1-i \) to \( 1+2i \) lies in the complex plane. In complex analysis, the techniques to solve such integrals can utilize various theorems and tools developed specifically for complex numbers.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a central theorem in calculus, and there is a version of this theorem suitable for complex functions and contour integrals. This theorem connects differentiation with integration and allows us to evaluate integrals conveniently.

• **In Complex Analysis:** If a function \( F(z) \) has a derivative \( f(z) \) at every point in a region, then \( F(z) \) is said to be an antiderivative of \( f(z) \).
• **Contour Integrals:** In such cases, the integral of \( f(z) \) along a curve from \( a \) to \( b \) can be calculated using the antiderivative: \[ \int_{a}^{b} f(z) dz = F(b) - F(a) \]

In the given exercise, the function \( F(z) = \frac{1}{2} e^{z^2} \) serves as an antiderivative of \( f(z) = z e^{z^2} \), and applying the Fundamental Theorem of Calculus for contour integrals, you find \( F(b) - F(a) \), where \( b = 1+2i \) and \( a = 1-i \).

Antiderivatives

An antiderivative of a function is a function whose derivative is the original function. Finding antiderivatives is crucial for evaluating integrals.

• **Identification:** To find an antiderivative, you need to recognize when the derivative of a function equals the given function. In the exercise, \( F(z) = \frac{1}{2} e^{z^2} \) is identified as the antiderivative of \( z e^{z^2} \).
• **Verification:** It is important to verify this by differentiating the proposed antiderivative and ensuring you arrive back at the original function: \( F'(z) = z e^{z^2} \).

Understanding and correctly identifying antiderivatives allow you to easily solve integrals using powerful theorems like the Fundamental Theorem of Calculus for contour integrals. Mastery of finding antiderivatives is key in both real and complex analysis.

Euler's Formula

Euler's formula connects complex exponentials with trigonometric functions. It is a remarkable discovery that aids significantly in solving expressions involving complex numbers.

• **The Formula:** Euler's formula states \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \). This is particularly useful as it relates exponential functions and trigonometry.
• **Application in Integration:** When working with complex numbers and contour integrals, Euler's formula is used to express complex exponentials in a form that is sometimes easier to interpret or calculate.

In the exercise, Euler's formula can help simplify expressions like \( e^{-3 + 4i} \) or \( e^{1 - 2i} \) into forms involving sine and cosine functions. Knowing how to use Euler's formula efficiently is extremely helpful in both theoretical and practical aspects of complex analysis.