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Magazine Chapter 6: Problem 18
Using partial fraction decomposition, show that if \(z\) lies in the right half- plane and \(C\) is the line segment joining 0 to \(z\), then $$ \int_{C} \frac{d \xi}{\xi^{2}+1}=\operatorname{Arctan} z=\frac{i}{2} \log (z+i)-\frac{i}{2} \log (z-i)+\frac{\pi}{2} $$
Short Answer
Expert verified
The integral equals \( \text{Arctan}(z) = \frac{i}{2}\log(z+i) - \frac{i}{2}\log(z-i) + \frac{\pi}{2} \).
Step by step solution
01
Express the Integral with Partial Fractions
Start by recognizing that the function \( \frac{1}{\xi^2 + 1} \) can be decomposed into partial fractions. The polynomial \( \xi^2 + 1 \) can be factored as \( (\xi + i)(\xi - i) \). Thus, we have the identity: \[ \frac{1}{\xi^2 + 1} = \frac{A}{\xi + i} + \frac{B}{\xi - i} \] To find \( A \) and \( B \), solve the equation \( 1 = A(\xi - i) + B(\xi + i) \).
02
Solve for Coefficients A and B
To solve for \( A \) and \( B \), substitute specific values of \( \xi \) to simplify the calculations:1. Set \( \xi = i \), then: \[ 1 = A(i - i) + B(i + i) = 2Bi \] This gives \( B = \frac{1}{2i} = -\frac{i}{2} \). 2. Set \( \xi = -i \), then: \[ 1 = A(-i + i) + B(-i - i) = -2Ai \] This gives \( A = -\frac{1}{2i} = \frac{i}{2} \).
03
Rewrite the Integral with Partial Fractions
With \( A \) and \( B \) found, rewrite the original integral using partial fractions:\[ \int_C \frac{d\xi}{\xi^2 + 1} = \int_C \left( \frac{i/2}{\xi + i} - \frac{i/2}{\xi - i} \right) d\xi \] This splits the integral into two separate integrals.
04
Integrate Each Partial Fraction Separately
Now integrate each part separately: 1. \( \int_C \frac{i/2}{\xi + i} d\xi = \frac{i}{2} \int_C \frac{1}{\xi + i} d\xi \) \[ \text{This is equal to } \frac{i}{2} \log(\xi + i) \] evaluated from 0 to \( z \). 2. \( \int_C -\frac{i/2}{\xi - i} d\xi = -\frac{i}{2} \int_C \frac{1}{\xi - i} d\xi \) \[ \text{This is equal to } -\frac{i}{2} \log(\xi - i) \] evaluated from 0 to \( z \).
05
Evaluate the Integrals
Substitute the limits of \( \xi \) as 0 to \( z \) for both integrals to get:\[\frac{i}{2} (\log(z + i) - \log(i)) - \frac{i}{2} (\log(z - i) - \log(-i))\] Simplify by combining the results (note that \( \log(-i) = \log(i) - i\pi \)) to complete the expression.
06
Simplify Using Properties of Logarithms
Combine the logarithms using the properties of logarithms:\[ \frac{i}{2} \log \left( \frac{z + i}{z - i} \right) + \frac{\pi}{2} \] Recognize this as equal to \( \text{Arctan}(z) \) based on its principal values and definition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
In complex analysis, Partial Fraction Decomposition is a technique used to express complex rational functions as a sum of simpler fractions. This method is particularly useful when dealing with integrals involving complex variables.
- Consider a rational function where the denominator can be factored into linear terms, such as \((\xi^2 + 1)\), which factors as \( (\xi + i)(\xi - i)\).
- The goal is to express the function \( \frac{1}{\xi^2 + 1}\) in terms of its partial fractions: \( \frac{A}{\xi + i} + \frac{B}{\xi - i}\).
Contour Integration
Contour integration involves evaluating integrals of complex functions along a curve (contour) in the complex plane. This powerful tool is extensively used in complex analysis, helping to solve many real-world problems involving complex variables.
- Here, our contour \(C\) is a straight line segment joining points 0 and \(z\) in the complex plane.
- The challenge is to integrate the function \( \frac{1}{\xi^2 + 1}\) using this contour.
- By breaking it down into simpler fractions using partial fraction decomposition, we can tackle the integral with more manageable parts.
Complex Logarithm
The complex logarithm extends the idea of a real logarithm to complex numbers. It is significant due to its involvement in expressing complex integrals. In the current problem, we encounter the complex logarithm while evaluating contour integrals.
- The complex logarithm function is defined as \(\log(z) = \ln|z| + i\arg(z)\), where \(|z|\) is the modulus and \(\arg(z)\) is the argument of the complex number.
- Since the logarithm is multi-valued, a branch cut is typically required, and a principal value is chosen, usually restricting \(\arg(z)\) to \([-\pi, \pi]\).
- In the context of this exercise, the logarithmic expressions \( \log(z+i) \) and \( \log(z-i) \) are critical. They describe how this function evolves along the contour from 0 to \(z\).
Arctangent Function
The arctangent function, noted as \( \operatorname{Arctan}(z)\), is the inverse tangent function applied to complex arguments. It provides a way to relate complex logarithms and angles.
- The arctangent function is understood as \(\operatorname{Arctan}(z) = \frac{i}{2} \log\left(\frac{z+i}{z-i}\right)\), which is evident in the last results of the solved problem.
- The function describes an angle whose tangent is \(z\), acting notably in the principal value between \(-\pi/2\) and \(\pi/2\).
- In the solution, the connection between the arctangent and contour integral simplifies the expression involving logarithms, leading finally to a straightforward result: \( \operatorname{Arctan}(z)\).
