Problem 22 - Advanced Engineering Mathematics

Question

We have \(\int_{C}\left(z^{2}-z+2\right) d z=\int_{C_{1}}\left(z^{2}-z+2\right) d z+\int_{C_{2}}\left(z^{2}-z+2\right) d z\) On \(C_{1}, y=1,0 \leq x \leq 1, z=x+i, d z=d x\), \(\int_{C_{1}}\left(z^{2}-z+2\right) d z=\int_{0}^{1}\left[(x+i)^{2}-x+2-i\right] d x=\frac{5}{6}\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}}\left(z^{2}-z+2\right) d z=i \int_{1}^{0}\left[(1+i y)^{2}+1-i y\right] d y=\frac{1}{2}-\frac{5}{3} i\). Thus \(\int_{C}\left(z^{2}-z+2\right) d z=\frac{1}{2}-\frac{5}{3} i+\frac{5}{6}=\frac{4}{3}-\frac{5}{3} i\).

Step-by-Step Solution

Answer

The integral is \(\frac{4}{3} - \frac{5}{3}i\).

Step 1: Break Down Integral Over Curve C

We start with the given statement that the integral over the curve C can be broken into two integrals over individual parts of the curve, C1 and C2: \[\int_{C}(z^2-z+2)\,dz = \int_{C_1}(z^2-z+2)\,dz + \int_{C_2}(z^2-z+2)\,dz\] Your problem also specifies that \(\int_{C_1}(z^2-z+2)\,dz = \frac{5}{6}\) and \(\int_{C_2}(z^2-z+2)\,dz = \frac{1}{2}-\frac{5}{3}i\).

Step 2: Substitute Integral Values

Substitute the given values for the integrals of \( C_1 \) and \( C_2 \) into the equation for the integral over the entire curve C:\[\int_{C}(z^2-z+2)\,dz = \frac{5}{6} + \left(\frac{1}{2} - \frac{5}{3}i\right)\]

Step 3: Simplify the Expression

To simplify, first convert the fractions to have a common denominator. The least common multiple of the denominators 6 and 2 is 6.Convert \(\frac{1}{2}\) to \(\frac{3}{6}\) since\[\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}\]Now combine the real parts of the expression:\[\frac{5}{6} + \frac{3}{6} = \frac{8}{6} = \frac{4}{3}\]

Step 4: Complete the Complex Number

Now, simply add the imaginary parts from the given integral over \( C_2 \) to get the final solution for the integral over curve C:The imaginary part is \(-\frac{5}{3}i\). Thus, combining the real and imaginary parts, we have:\[\int_{C}(z^2-z+2)\,dz = \frac{4}{3} - \frac{5}{3}i\]

Step 5: Conclusion

We have calculated the integral over the entire curve C by summing the integrals over individual sections \( C_1 \) and \( C_2 \), providing us with a final result \[\frac{4}{3} - \frac{5}{3}i\] for the integral.

Key Concepts

Curve Integrals

Curve integrals, also known as line integrals, are a fundamental concept in complex analysis and vector calculus. They allow us to integrate complex-valued functions over a path in the complex plane. This is especially useful for functions that are difficult or impossible to integrate over standard intervals.
When performing a curve integral, one must choose a path or curve along which to integrate the function. In this case, the curves \(C_1\) and \(C_2\) are chosen as paths.

• For the curve \(C_1\), the path is a horizontal line \(y=1\) as \(x\) ranges from 0 to 1.
• For the curve \(C_2\), the path is a vertical line \(x=1\) as \(y\) ranges from 0 to 1.

The curve integral of a function \(f(z)\) along a path \(C\) is generally represented by \(\int_C f(z) \, dz\). It's crucial to split the integral into parts if different path segments are involved, as shown in the original exercise. By evaluating each segment separately and summing their results, we find the integral over the entire path \(C\). This ability to break down complex paths into simpler segments makes curve integrals a powerful tool in complex analysis.

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are usually written in the form \(z = x + yi\), where \(x\) is the real part and \(iy\) is the imaginary part. In the case of complex integration, understanding complex numbers aids in analyzing and performing calculations effectively.
Each complex number can be plotted in a coordinate system known as the complex plane.

• The real part \(x\) determines the position along the horizontal axis.
• The imaginary part \(yi\) determines the position along the vertical axis.

From the example provided, \(z = x + i\) on \(C_1\) and \(z = 1 + iy\) on \(C_2\) are specific representations of complex numbers along their respective paths. By substituting these values into the function and evaluating the integral, one effectively navigates both the real and imaginary components of the curve. The result is often written as \(a + bi\), where \(a\) represents the real part and \(bi\) represents the imaginary part of the outcome. Complex numbers thus allow us to compactly perform arithmetic on two-dimensional numbers, making them ideal for handling multi-part paths and functions in complex integration.

Analytic Functions

Analytic functions play a prominent role in complex analysis. They are functions that are locally defined by a convergent power series. In simpler terms, an analytic function is one that is differentiable at every point in its domain. This differentiability extends into the complex plane, which provides a broad framework for advanced mathematical operations.
The function \(f(z) = z^2 - z + 2\) from our exercise is an example of an analytic function. It is a polynomial, and polynomials are always analytic in the entire complex plane. This property makes them very suitable for complex integration tasks.
Analytic functions have several vital properties:

• They can be expressed as a power series around any point in their domain.
• Their behavior is well understood, leading to stable integration over complex paths.
• Being infinitely differentiable, they conform well to powerful theories like Cauchy's Integral Theorem.

In the process of complex integration, the function's analyticity ensures that integrals can be evaluated effectively over well-defined paths. This means that the curve integrals engage both their real and imaginary parts seamlessly, providing robust mathematical frameworks for complex solutions.