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Chapter 18: Problem 12

\(\int_{C} \sin z d z=\int_{C_{1}} \sin z d z+\int_{C_{2}} \sin z d z\) where \(C_{1}\) and \(C_{2}\) are the line segments \(y=0,0 \leq x \leq 1,\) and \(x=1\), \(0 \leq y \leq 1,\) respectively. Now \(\int_{C_{1}} \sin z d z=\int_{0}^{1} \sin x d x=1-\cos 1\) \(\int_{C_{2}} \sin z d z=i \int_{0}^{1} \sin (1+i y) d y=\cos 1-\cos (1+i)\). Thus \(\int_{C} \sin z d z=(1-\cos 1)+(\cos 1-\cos (1+i))=1-\cos (1+i)=(1-\cos 1 \cosh 1)+i \sin 1 \sinh 1=0.1663+0.9889 i\).

Short Answer

Expert verified
The integral is \(0.1663 + 0.9889i\).

Step by step solution

01

Understanding the Problem

We are given an integral problem on a contour composed of two line segments, \(C_1\) and \(C_2\), and need to calculate \(\int_C \sin z \, dz\). The path \(C\) is divided into two segments: \(C_1\) is a line segment on the real axis from 0 to 1, and \(C_2\) is a vertical line segment from (1,0) to (1,1).
02

Calculating \(\int_{C_1} \sin z \ dz\)

\(C_1\) corresponds to the real line segment from 0 to 1. Therefore, \(z = x\) where \(y = 0\). The integral thus simplifies to \(\int_{0}^{1} \sin x \, dx\). Solving this gives \(1 - \cos 1\).
03

Parameterizing and Calculating \(\int_{C_2} \sin z \ dz\)

\(C_2\) is a vertical segment, thus \(z = 1 + iy\) where \(x=1\) remains constant and \(0 \leq y \leq 1\). Thus, \(dz = i \, dy\). The integral becomes \(i \int_{0}^{1} \sin(1+iy) \, dy\). Breaking it down, it evaluates to \(\cos 1 - \cos(1+i)\).
04

Summing the Integrals

Summing the results from \(\int_{C_1} \sin z \ dz\) and \(\int_{C_2} \sin z \ dz\), we have \((1 - \cos 1) + (\cos 1 - \cos(1+i))\). This simplifies to \(1 - \cos(1+i)\).
05

Evaluating \(1 - \cos(1+i)\)

To evaluate, we use the identity \(\cos(a+bi) = \cos(a) \cosh(b) - i \sin(a) \sinh(b)\). Applying it to \(\cos(1+i)\), we get \(\cos 1 \cosh 1 - i \sin 1 \sinh 1\). Therefore, \(1 - \cos(1+i) = (1 - \cos 1 \cosh 1) + i \sin 1 \sinh 1\).
06

Final Numerical Evaluation

Calculating the above terms numerically, we find \(1 - \cos 1 \cosh 1 = 0.1663\) and \(\sin 1 \sinh 1 = 0.9889\). Thus the final result is \(0.1663 + 0.9889i\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Contour Integration
Contour integration is a powerful technique in complex analysis used to evaluate certain types of integrals over a path in the complex plane. One common application is integrating complex functions along specified paths called contours. In this particular exercise, the path is composed of two segments, noted as \( C_1 \) and \( C_2 \). These segments outline a route that forms a contour for integration.
  • Path Definition: The contour \( C \) consists of a horizontal segment from \( 0 \) to \( 1 \) (\( C_1 \)), and a vertical segment from \( 1 \) to \( 1 + i \) (\( C_2 \)).
  • Complex Function: We are dealing with the function \( \sin z \), where \( z \) is a complex number.
  • Segment Calculation: Each segment of the contour is evaluated separately before summing the results to solve the integral over the entire contour.
This method leverages the properties of analytic functions, making contour integration a cornerstone of complex analysis.
Line Integrals
Line integrals extend the concept of integration by taking into account a path through a region. In the real line, integration is performed over intervals, while in complex analysis, line integrals follow curves in the complex plane. Here, the integral evaluates \( \sin z \) along two different line segments which contribute to the overall value.The integration over the line segments proceeds as follows:
  • For \( C_1 \): Because this segment lies along the real axis (where \( y = 0 \)), the integral simplifies to \( \int_{0}^{1} \sin x \, dx \), resulting in \( 1 - \cos 1 \).
  • For \( C_2 \): This involves a vertical path where \( x = 1 \) and \( z = 1 + iy \), which requires additional steps to solve. The transformation of \( dz = i \, dy \) and integration leads to the result \( \cos 1 - \cos(1+i) \).
Line integrals allow us to evaluate functions over paths within the complex plane, taking into account both the curve's geometry and the function's behavior along that path.
Complex Analysis
Complex analysis is the branch of mathematics that investigates functions of complex numbers. It presents unique challenges and results, which are both intriguing and powerful in mathematical applications. In the scenario discussed, the concept of complex analysis comes into play when dealing with the integral of \( \sin z \) along complex paths.Key aspects explored in this example include:
  • Complex Functions: The function \( \sin z \) is an extension of the sine function to the complex plane, implying it can take complex arguments.
  • Analyticity: Functions like \( \sin z \) that are analytic (holomorphic) allow us to apply methods of complex integration because they exhibit nice properties such as being differentiable everywhere in their domain.
  • Evaluation Techniques: Splitting the contour into segments and handling each separately makes computation more manageable. Each portion of the contour uses complex numbers specific to that segment's definition.
Complex analysis enables powerful integration techniques like contour integration, which offer elegant solutions to integrals involving complex functions.
Exponential Functions
Exponential functions play a crucial role in complex analysis. They are often encountered when evaluating integrals involving trigonometric functions over complex paths. When complex numbers are involved, sinusoidal functions are expressed using exponential functions through Euler鈥檚 formula.In this case:
  • The expression \( \cos(1+i) \) is broken into its real and imaginary parts using the identity:
  • \[ \cos(a + bi) = \cos(a) \cosh(b) - i \sin(a) \sinh(b) \]
These identities help simplify and solve expressions that arise in complex integration. Here, it allows us to represent \( \cos(1+i) \) in terms of real functions (hyperbolic cosine and sine functions), which are more straightforward to evaluate.Understanding how exponential functions underpin complex expressions aids significantly in tackling line integrals involving trigonometric functions like \( \sin z \). By breaking down the problem, students gain clarity on how to manage such calculations effectively.

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Most popular questions from this chapter

$$\begin{aligned} \int_{1-2 i}^{\pi i} \cos z d z &=\left.\sin z\right|_{1-2 i} ^{\pi i}=\sin \pi i-\sin (1-2 i)=i \sinh \pi-[\sinh 1 \cosh 2-i \cos 1 \sinh 2] \\ &=-\sin 1 \cosh 2+i(\sinh \pi+\cos 1 \sinh 2)=-3.1658+13.5083 i \end{aligned}$$

Using \(z=e^{i t}=\cos t+i \sin t, d z=(-\sin t+i \cos t) d t\) and \(x=\cos t\), \(\begin{aligned} \oint_{C} \operatorname{Re}(z) d z &=\int_{0}^{2 \pi} \cos t(-\sin t+i \cos t) d t=-\int_{0}^{2 \pi} \sin t \cos t d t+i \int_{0}^{2 \pi} \cos ^{2} t d t \\ &=-\frac{1}{2} \int_{0}^{2 \pi} \sin 2 t d t+\frac{1}{2} i \int_{0}^{2 \pi}(1+\cos 2 t) d t=\pi i \end{aligned}\).

(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) \(\int_{C} e^{z} d z=\int_{0}^{3+i} e^{z} d z=\left.e^{z}\right|_{0} ^{3+i}=e^{3+i}-e^{0}=\left(e^{3} \cos 1-1\right)+i e^{3} \sin 1\)

By partial fractions, \(\oint_{C} \frac{2 z+1}{z(z+1)} d z=\oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z\) (a) By Theorem 18.4 and (4) of Section 18.2 $$\oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z=2 \pi i+0=2 \pi i$$ (b) \(\quad\) By writing \(\oint_{C}=\oint_{C_{1}}+\oint_{C_{2}}\) where \(C_{1}\) and \(C_{2}\) are the circles \(|z|=1 / 2\) and \(|z+1|=1 / 2,\) respectively we have by Theorem 18.4 and (4) of Section 18.2 $$\begin{aligned} \oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z &=\oint_{C_{1}} \frac{1}{z} d z+\oint_{C_{1}} \frac{1}{z+1} d z+\oint_{C_{2}} \frac{1}{z} d z+\oint_{C_{2}} \frac{1}{z+1} d z \\ &=2 \pi i+0+0+2 \pi i=4 \pi i \end{aligned}$$(c) \(\operatorname{since} f(z)=\frac{2 z+1}{z(z+1)}\) is analytic within and on \(C\) it follows from Theorem 18.4 that $$\oint_{C} \frac{2 z+1}{z^{2}+z} d z=0$$

We have \(\oint_{C} x d z=\int_{C_{1}} x d z+\int_{C_{2}} x d z+\int_{C_{3}} x d z\) On \(C_{1}, y=0,0 \leq x \leq 1, z=x, d z=d x\), \(\int_{C_{1}} x d z=\int_{0}^{1} x d x=\frac{1}{2}\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}} x d z=i \int_{0}^{1} d y=i\). On \(C_{3}, y=x, 0 \leq x \leq 1, z=x+i x, d z=(1+i) d x\), \(\int_{C_{3}} x d z=(1+i) \int_{1}^{0} x d x=-\frac{1}{2}-\frac{1}{2} i\). Thus \(\oint_{C} x d z=\frac{1}{2}+i-\frac{1}{2}-\frac{1}{2} i=\frac{1}{2} i\).
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