Problem 13 - Advanced Engineering Mathematics

Question

We have \(\quad \int_{C} \operatorname{Im}(z-i) d z=\int_{C_{1}}(y-1) d z+\int_{C_{2}}(y-1) d z\) On \(C_{1}, z=e^{i t}, 0 \leq t \leq \pi / 2, d z=i e^{i t} d t, y=\sin t=\left(e^{i t}-e^{-i t}\right) / 2 i\), \(\int_{C_{1}}=(y-1) d z=\frac{1}{2} \int_{0}^{\pi / 2}\left[e^{i t}-e^{-i t}-2 i\right] e^{i t} d t=\frac{1}{2} \int_{0}^{\pi / 2}\left[e^{2 i t}-1+2 i e^{i t}\right] d t=1-\frac{\pi}{4}-\frac{1}{2} i\).On \(C_{2}, y=x+1,-1 \leq x \leq 0, z=x+(x+1) i, d z=(1+i) d x\), \(\int_{C_{2}}(y-1) d z=(1+i) \int_{0}^{-1} x d x=\frac{1}{2}+\frac{1}{2} i\). Thus \(\int_{C} \operatorname{Im}(z-i) d z=\left(1-\frac{\pi}{4}-\frac{1}{2} i\right)+\left(\frac{1}{2}+\frac{1}{2} i\right)=\frac{3}{2}-\frac{\pi}{4}\).

Step-by-Step Solution

Answer

\(\int_{C} \operatorname{Im}(z-i) \, dz = \frac{3}{2} - \frac{\pi}{4}\).

Step 1: Understand the Contour Integral

We need to evaluate the integral \( \int_{C} \operatorname{Im}(z-i) \, dz \) which is split into two parts: \( \int_{C_{1}}(y-1) \, dz \) and \( \int_{C_{2}}(y-1) \, dz \). Each part represents an integral along a different path portion, \(C_{1}\) and \(C_{2}\).

Step 2: Evaluate the Integral Along Path C1

For the path \(C_{1}\), we have \(z = e^{it} \) with \(0 \leq t \leq \frac{\pi}{2}\). Hence, \(dz = ie^{it} dt \) and \(y = \sin t = \frac{e^{it} - e^{-it}}{2i}\). The integral becomes: \[\int_{C_{1}}(y-1) dz = \frac{1}{2} \int_{0}^{\pi/2} \left[ e^{2it} - 1 + 2i e^{it} \right] \, dt\]After evaluating, it simplifies to: \[1 - \frac{\pi}{4} - \frac{1}{2} i\]

Step 3: Evaluate the Integral Along Path C2

For the path \(C_{2}\), we have \( y = x + 1\) with \(-1 \leq x \leq 0\), and \(z = x + (x+1)i\). Thus, \(dz = (1+i) dx\). The integral along \(C_{2}\) is:\[\int_{C_{2}}(y-1) dz = (1+i) \int_{0}^{-1} x \, dx\]After evaluating it simplifies to:\[\frac{1}{2} + \frac{1}{2} i\]

Step 4: Combine the Integrals

To find the total integral \(\int_{C} \operatorname{Im}(z-i) \, dz\), we sum the results from \(C_{1}\) and \(C_{2}\):\[\left(1 - \frac{\pi}{4} - \frac{1}{2} i\right) + \left(\frac{1}{2} + \frac{1}{2} i\right) = \frac{3}{2} - \frac{\pi}{4}\]

Key Concepts

Complex Analysis

Complex analysis is a significant branch of mathematics focusing on functions of complex numbers. It explores concepts such as continuity, differentiability, and integrability of complex functions. This branch is pivotal for understanding phenomena like contour integrals, which involve integrating along paths in the complex plane.
In the context of this topic, complex analysis provides tools for analyzing functions along paths, as seen in the \( \int_{C} \operatorname{Im}(z-i) \, dz\) integral. Here, complex functions guide how integrals behave over specific contours, enabling deeper insights into the properties of varied complex functions.
Complex analysis leverages the powerful tool of contour integration, offering a way to traverse within the scope of the complex plane.

• Functions familiar in real analysis can have overlapping principles in the complex plane, allowing complex analysis to expand their utility.
• Tools from complex analysis, such as the Fundamental Theorem of Calculus and Cauchy’s Integral Theorem, play a critical role in solving such integrals.

Contour Path

The contour path refers to the specific trajectory along which integration occurs in the complex plane. Understanding these paths is vital for evaluating integrals, like the ones found in contour integrals.
In our exercise, we have two paths, \(C_1\) and \(C_2\), each designed to traverse part of the complex plane. For \(C_1\), the path is defined using the parameter \(t\), representing an arc of a circle, while \(C_2\) is a straight line segment.
Choosing an appropriate contour path is crucial as it affects the integral’s outcome and simplifies the calculation process.

• Contours may be smooth paths or piecewise smooth, meaning they can include bends and corners.
• Understanding the nature of the contour can aid in applying relevant theorems effectively, especially those that hinge upon the path's topology and properties.

Integral Calculation

Integral calculation is fundamental to our exercise, which involves determining the value of an integral along a specified contour path. This process encompasses several steps:
First, break down the overall contour into individual segments, such as \(C_1\) and \(C_2\). Each integral is evaluated separately before combining their results.
For \(C_1\), the integration parameter \(t\) defines the path, and substituting \(e^{it}\) and \(dz = ie^{it} dt\) allows the integral \( \int_{0}^{\pi/2} [e^{2it} - 1 + 2i e^{it}] \, dt\) \( to be evaluated elementarily.\)
Similarly, for \(C_2\), \(dz = (1+i) dx\) simplifies integration over the given path.

• Breaking down complex integrals into manageable pieces makes them easier to handle.
• Properly parameterizing contours aids in converting path integrals into real integrals.
• Summing smaller, individually evaluated integrals provides the total integral value.

Complex Plane

The complex plane is a crucial framework in complex analysis, serving as a visual and conceptual representation of complex numbers. Each point on the plane corresponds to a complex number \(z = x + yi\), where \(x\) is the real part, and \(y\) is the imaginary part.
In our scenario, the paths \(C_1\) and \(C_2\) navigate through this plane. The characteristic pathways in this plane, whether circular arcs or line segments, underpin calculations like contour integrations. The exercise in question exemplifies how specific complex number paths manifest operations over the complex plane.

• The horizontal axis (real) and vertical axis (imaginary) help distinguish complex numbers' components.
• Understanding the complex plane is vital for visualizing and solving problems involving complex functions and integrals.