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Magazine Chapter 19: Problem 27
Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. \(\oint_{C} \frac{\tan z}{z} d z, C:|z-1|=2\)
Short Answer
Expert verified
The integral evaluates to \( 2\pi i \).
Step by step solution
01
Identify the singularities
First, identify the singularities of the function \( f(z) = \frac{\tan z}{z} \) inside the contour \( C: |z-1|=2 \). The tangent function, \( \tan z = \frac{\sin z}{\cos z} \), has poles where \( \cos z = 0 \), which occur at \( z = \frac{(2n+1)\pi}{2} \), where \( n \) is an integer. Additionally, there is a pole at \( z = 0 \) because of the \( \frac{1}{z} \) term. Check which of these are inside the contour \( |z-1| = 2 \).
02
Analyze location of singularities relative to contour
The contour \( |z-1| = 2 \) is centered at \( z = 1 \) with radius 2. For the poles of \( \tan z \), convert \( z = \frac{(2n+1)\pi}{2} \) into an approximate numerical value to determine whether they lie within this contour. We need singularities such as \( z = \frac{\pi}{2} \approx 1.57 \) and \( z = \ rac{3\pi}{2} \approx 4.71 \). However, only \( z = \frac{\pi}{2} \) and \( z = 0 \) (since \( |0-1| < 2 \) ) lie inside this contour.
03
Calculate the residues
Calculate the residues at the singularities inside the contour: \( z = 0 \) and \( z = \frac{\pi}{2} \). For \( z = 0 \), the residue can be found by multiplying through by \( z \) and taking the limit as \( z \to 0 \), resulting in \( \lim_{z \to 0} \tan z = 0 \). For \( z = \frac{\pi}{2} \), since it is a simple pole of \( \tan z \), compute the residue using the formula for poles \( \text{Residue of } \tan z = \lim_{z \to \frac{\pi}{2}} (z - \frac{\pi}{2})\tan z = 1 \). In fact, due to symmetry and periodicity, the residue of \( \tan z \) at any \( \frac{(2n+1)\pi}{2} \) is 1.
04
Apply Cauchy's residue theorem
Apply Cauchy's residue theorem: \( \oint_{C} f(z) \, dz = 2\pi i \times \sum \text{Residues of } f(z) \text{ inside } C \). The sum of residues inside the contour \( C \) is \( 0 + 1 = 1 \). Thus, the integral evaluates to \( 2\pi i \times 1 = 2\pi i \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Analysis
Complex analysis is a branch of mathematics that deals with functions of a complex variable. It combines real and imaginary numbers to analyze the behavior and properties of complex functions. A complex number is expressed in the form \( z = x + yi \), where \( x \) and \( y \) are real numbers and \( i \) is the imaginary unit, defined by \( i^2 = -1 \).
- Complex functions are functions that take complex numbers as inputs and output complex numbers.
- The study of these functions includes concepts like limits, continuity, and differentiation.
- Complex analysis also explores the unique properties of analytic functions, which are functions that are differentiable at every point in their domain.
Contour Integration
Contour integration is a method used in complex analysis to evaluate integrals of complex functions along a specific path鈥攐r contour鈥攊n the complex plane. This technique is particularly useful for functions with singularities, as it allows the integral to be calculated without actually integrating over the singularities themselves.
- A contour is any continuous curve in the complex plane.
- Choosing the appropriate contour is key to simplifying the calculation of complex integrals.
- Contour integrals can be closed, meaning they form a loop that returns to the starting point.
Singularities
Singularities are points at which a complex function is not defined or not analytic. In the context of functions of a complex variable, singularities take on special importance, as they often determine the behavior of a function around those points.
- Poles are a common type of singularity, where a function approaches infinity.
- Essential singularities are points where the behavior of the function is more erratic, with no limit as you approach the point.
- Branch points are where a function cannot return to a single value, leading to multi-valued results.
Residue Calculation
Residue calculation involves determining the residues of a function at its singularities. This is an essential step in applying Cauchy's residue theorem for evaluating complex integrals.
- The residue of a function at a pole is the coefficient of \( \frac{1}{z-a} \) when the function is expressed in its Laurent series around the pole \( a \).
- For simple poles, the residue can be found directly by taking limits or using residue formulas.
- Residue calculation simplifies the evaluation of contour integrals by reducing the problem to finding sums of residues inside the contour.
