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Magazine Chapter 19: Problem 3
Use a Laurent series to find the indicated residue. \(f(z)=\frac{4 z-6}{z(2-z)} ; \operatorname{Res}(f(z), 0)\)
Short Answer
Expert verified
The residue of \( f(z) \) at \( z = 0 \) is \(-3\).
Step by step solution
01
Identify the Singularities
The function is given by the fraction \( f(z) = \frac{4z - 6}{z(2-z)} \). The singularities occur where the denominator is zero, that is, \( z = 0 \) and \( z = 2 \). For this problem, we are interested in the singularity at \( z = 0 \).
02
Expand Denominator around Singularity
Express \( \frac{1}{2-z} \) using a geometric series expansion around \( z=0 \):\[\frac{1}{2-z} = \frac{1}{2\left(1 - \frac{z}{2}\right)} = \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\]This expansion is valid for \( |z| < 2 \).
03
Derive Laurent Series for \( f(z) \)
Substitute the expanded form of \( \frac{1}{2-z} \) into \( f(z) \):\[f(z) = \frac{4z-6}{z} \cdot \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n = \left(4 - \frac{6}{z}\right) \cdot \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\]Simplify this to get the Laurent series:\[2 \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n - 3 \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n \cdot \frac{1}{z}\]Which simplifies further to:\[\sum_{n=0}^{\infty} 2 \left(\frac{z}{2}\right)^n - \sum_{n=0}^{\infty} 3 \frac{z^n}{2^n z}\]
04
Solve for the Residue
The residue is the coefficient of the \( \frac{1}{z} \) term in the Laurent series. In our expanded series:\[-\sum_{n=0}^{\infty} 3 \frac{z^n}{2^n z} = -3\]Thus, the residue of \( f(z) \) at \( z = 0 \) is \( -3 \).
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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 fascinating branch of mathematics that studies complex numbers and the functions that depend on them. Here are a few key ideas:
- Complex numbers are numbers of the form \(a + bi\), where \(i\) is the imaginary unit, equal to the square root of \(-1\).
- Functions of complex variables can exhibit behaviors quite different than their real counterparts.
- Complex analysis includes many powerful techniques, such as the use of complex integrals, series expansions, and analytic continuation.
Singularities
A singularity in complex analysis is a point where a function ceases to be analytic or well-behaved. For our function \(f(z) = \frac{4z - 6}{z(2-z)}\), singularities are where the denominator equals zero:
- \(z = 0\) and \(z = 2\) are the singular points of the function.
- At these points, the function is undefined or infinite.
- Singularities can be classified as removable, poles, or essential depending on their nature.
Residue Theorem
The residue theorem is a central result in complex analysis used for evaluating integrals of complex functions around closed contours. Here's a quick summary:
- The residue of a function at a singular point is the coefficient of the \(\frac{1}{z}\) term in its Laurent series expansion about that point.
- The residue theorem allows for transformation of contour integrals of a function into a summation of residues inside the contour.
- This theorem simplifies complex integral calculations significantly.
Geometric Series
A geometric series is a sum of terms, each which is a fixed multiple of the previous one. In mathematical terms, a geometric series with a common ratio \(r\) and first term \(a\) is described as:
- \(a + ar + ar^2 + ar^3 + \ldots\)
- The series converges if \(|r| < 1\).
- Summing an infinite geometric series when it converges is done using the formula: \(\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\).
