Problem 8 - Complex Analysis for Mathematics and Engineering

Question

Let \(f\) be analytic and have a zero of order \(k\) at \(z_{0} .\) Show that the funetion \(\frac{1}{f}\) has a simple pole at zo.

Step-by-Step Solution

Answer

\( \frac{1}{f} \) has a simple pole at \( z_0 \) if \( f \) has a zero of order 1 there.

Step 1: Understanding the Problem

We are given that the function \( f \) is analytic and has a zero of order \( k \) at \( z_0 \). This means that we can express \( f(z) \) near \( z_0 \) as \( f(z) = (z-z_0)^k g(z) \), where \( g(z) \) is analytic and \( g(z_0) eq 0 \). We need to show that \( \frac{1}{f} \) has a simple pole at \( z_0 \).

Step 2: Expressing the Reciprocal Function

Since \( f(z) = (z - z_0)^k g(z) \), the reciprocal function \( \frac{1}{f(z)} \) can be written as:\[\frac{1}{f(z)} = \frac{1}{(z - z_0)^k g(z)}.\]

Step 3: Analyzing the Reciprocal for a Pole

We know that \( g(z) \) is analytic and nonzero at \( z_0 \). The function \( \frac{1}{g(z)} \) is therefore analytic at \( z_0 \). The expression \( \frac{1}{(z-z_0)^k} \) behaves as \( (z-z_0)^{-k} \).

Step 4: Determining the Type of Pole

The term \( (z-z_0)^{-k} \) in \( \frac{1}{f(z)} \) has its lowest power as \( -k \). For \( \frac{1}{f} \) to have a pole of any order, one power in the denominator must cancel out analytically. Therefore, the presence of \( (z-z_0)^{-1} \) at leading term confirms that \( \frac{1}{f} \) has a simple pole at \( z_0 \).

Step 5: Concluding the Pole Order

Since \( f(z) \) had a zero of order \( k \) and \( \frac{1}{f} \) is \( (z-z_0)^{-k} \cdot \frac{1}{g(z)} \), the simple pole exists when \( k = 1 \). A zero of order \( 1 \) at \( z_0 \) ensures that \( \frac{1}{f} \) has a simple pole at \( z_0 \).

Key Concepts

Analytic Functions

An analytic function is a fundamental concept in complex analysis. These functions are complex-valued and can be represented by a power series within their domain of convergence. This means they are infinitely differentiable, and their behavior is smooth and predictable.

To consider a function analytic, it must be complex differentiable at each point within an open subset of the complex plane. This differs from real functions, where differentiation might only apply at isolated points or intervals. Analytic functions capture the ideas of limits and continuity from calculus, but in the more generalized field of complex numbers.

Analytic functions exhibit various properties:

• They can be expanded into a Taylor series about any point where they are analytic.
• The radius of convergence in their power series is always positive, meaning they extend over some neighborhood of each point.
• Properties like Cauchy-Riemann conditions must be fulfilled, connecting these functions to harmonic potentials.

Understanding analytic functions is crucial for exploring concepts like zeros and poles within complex analysis.

Zero of a Function

The zero of a function, especially in complex analysis, refers to a point where a complex function equals zero. It's similar to finding roots of polynomial equations in algebra.

In the context of analytic functions, zeros can have various orders:

• If a zero has order 1, it is called a simple zero.
• When the zero's order is higher than 1, it indicates a repeated root.

The order of a zero of a function at a point such as \( z_0 \) is the smallest positive integer \( k \) for which the nth derivative at \( z_0 \) vanishes, indicating how many times the function \( f(z) \) touches the zero at that point.

In mathematical terms, if \( f(z) \) has a zero of order \( k \) at \( z_0 \), then it can be expressed as:\[ f(z) = (z - z_0)^k g(z) \] where \( g(z) \) is analytic and non-zero at \( z_0 \). This decomposition into a simpler form aids in analyzing the behavior of \( f(z) \) near its zeros.

Simple Pole

A simple pole is a type of singularity in complex analysis where a function behaves like \( \frac{1}{z-z_0} \) close to a point \( z_0 \). It is called "simple" because it has a multiplicity of one, meaning it doesn't repeat or extend into higher powers.

When analyzing the reciprocal of an analytic function, like \( \frac{1}{f} \), poles arise where the original function \( f \) has zeros:

• If \( f(z) \) has a zero of order 1 at \( z_0 \), \( \frac{1}{f} \) will have a simple pole at the same point.
• A simple pole implies that the Laurent series of the function at that point has just a single negative power term, namely \( (z-z_0)^{-1} \).

Understanding poles is essential because they affect the contour integration in the complex plane, providing insights into residues and complex function behavior.