Chapter 8: Problem 3
Let the function \(f\) be holomorphic in an open set containing the point \(z_{0}\) and have a zero at \(z_{0}\) of order \(m\). Prove that $$ \operatorname{res}_{z_{0}} \frac{f^{\prime}}{f}=m. $$
Short Answer
Expert verified
The residue is \(m\).
Step by step solution
01
Understand the Residue
The residue of a function at a certain point is the coefficient of \(\frac{1}{z-z_0}\) when the function is expressed as a Laurent series around the point \(z_0\). In this problem, we need to find the residue of \(\frac{f'}{f}\) at \(z_0\), given that \(f\) has a zero of order \(m\) at \(z_0\).
02
Write the Function at Zero
Since \(f\) is holomorphic and has a zero of order \(m\) at \(z_0\), it can be written as \(f(z) = (z-z_0)^m g(z)\), where \(g(z)\) is holomorphic and \(g(z_0) eq 0\).
03
Differentiate the Function
Differentiate \(f(z) = (z-z_0)^m g(z)\) using the product rule:\[f'(z) = m(z-z_0)^{m-1}g(z) + (z-z_0)^m g'(z).\]
04
Express the Quotient
Plug \(f(z)\) and \(f'(z)\) into \(\frac{f'}{f}\):\[\frac{f'(z)}{f(z)} = \frac{m(z-z_0)^{m-1}g(z) + (z-z_0)^m g'(z)}{(z-z_0)^m g(z)}.\]
05
Simplify the Expression
Divide each term by \((z-z_0)^m g(z)\):\[= \frac{m(z-z_0)^{m-1}g(z)}{(z-z_0)^m g(z)} + \frac{(z-z_0)^m g'(z)}{(z-z_0)^m g(z)}\]\[= \frac{m}{z-z_0} + \frac{g'(z)}{g(z)}.\]
06
Identify the Residue
The Laurent series expression \(\frac{m}{z-z_0} + \frac{g'(z)}{g(z)}\) shows that the coefficient of \(\frac{1}{z-z_0}\) is \(m\), thus the residue is \(m\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laurent series
The Laurent series is a powerful tool in complex analysis. It provides a way to represent a complex function as an infinite sum of terms, which include both positive and negative powers of (z - z_0). This is particularly useful near singular points, where the function might not be analytically expressible otherwise. The general form of a Laurent series of a function f(z) around a point z_0 is:\[f(z) = \sum_{{n=-\infty}}^{{\infty}} a_n (z - z_0)^n\]
- The terms where n is negative represent the principal part of the series.
- The coefficient a_{-1} is called the residue and is crucial in evaluating contour integrals using the residue theorem.
holomorphic functions
Holomorphic functions are the backbone of complex analysis. A function f(z) is said to be holomorphic at a point z_0 if it is complex differentiable in a neighborhood around z_0. This differentiability implies that f(z) is smooth and has a beautifully structured infinite Taylor series expansion. Mathematically, we define it as:\[f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n\]
- Being holomorphic means the function is not only continuous but infinitely differentiable within its radius of convergence.
- Holomorphic functions exhibit pleasant properties such as the ability to be represented locally by power series.
- If f is holomorphic at z_0, it ensures that not only is differentiation valid, but also operations like integration around closed contours are feasible and smooth.
order of a zero
Understanding the order of a zero is key when dealing with complex functions. If a function f(z) has a zero of order m at z_0, it means f(z) can be expressed as:\[f(z) = (z - z_0)^m g(z)\]where g(z) is holomorphic and non-zero at z_0. This concept is very important because:
- The order m indicates how the function behaves near z_0.
- It governs how many derivatives of f vanish at z_0 and affects the expansion of f(z).
- The order of zero directly influences the residue at that point when evaluating limits and integrals.