Problem 21 - A First Course in Complex Analysis with Applications

Question

Determine the order of the poles for the given function. $$ f(z)=\frac{1-\cosh z}{z^{4}} $$

Step-by-Step Solution

Answer

The function has a pole of order 2 at \(z = 0\).

Step 1: Identify the form of the function

We are given the function \( f(z) = \frac{1 - \cosh z}{z^4} \). We need to analyze the nature of \(1 - \cosh z\) around the origin, as this will help determine the poles and their orders.

Step 2: Expand \(\cosh z\) using Taylor series

The Taylor series expansion of \(\cosh z\) around \(z = 0\) is given by: \[\cosh z = 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \frac{z^6}{6!} + \cdots\].Thus, \(1 - \cosh z\) near \(z = 0\) becomes: \[1 - \cosh z = -(\frac{z^2}{2} + \frac{z^4}{24} + \cdots)\].This expression reveals the behavior of \(f(z)\) near zero.

Step 3: Simplify \(1-\cosh z\) for pole identification

From Step 2, we have \[1 - \cosh z = -\frac{z^2}{2} - \frac{z^4}{24} + \cdots\]. Near \(z = 0\), the behavior is dominated by the term \(-\frac{z^2}{2}\). Therefore, we can approximate \[1 - \cosh z \approx -\frac{z^2}{2}\] for small \(z\).

Step 4: Determine the behavior of \(f(z)\) around zero

Substitute the approximation from Step 3 into \(f(z)\):\[f(z) = \frac{-\frac{z^2}{2}}{z^4} = -\frac{1}{2z^2}\].This expression indicates a pole of order 2 at \(z = 0\), since the term in the denominator cancels exactly 2 powers of \(z\) in the numerator.

Key Concepts

Taylor Series

The Taylor series is an essential concept in complex analysis and calculus. It allows us to represent a function as an infinite sum of terms calculated from the derivatives of the function at a single point. More precisely, for a function \( f(z) \) that is infinitely differentiable at \( z = a \), the Taylor series expansion around this point is:

• \( f(z) = f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \frac{f'''(a)}{3!}(z-a)^3 + \cdots \)

Using Taylor series can simplify complex functions significantly, especially when analyzing their behavior near a specific point. In the context of our exercise, the Taylor series was used to expand \( \cosh z \) about \( z = 0 \). This method helps us to realize that \( 1 - \cosh z \) behaves as \( -\frac{z^2}{2} \) when \( z \) is close to zero. This expansion is crucial for determining the nature of poles in the function \( f(z) \).

Poles of a Function

Poles are singular points of a complex function where the function goes to infinity. They are characterized by the order, which tells us about the behavior and nature of the function near that point. In mathematical terms, for a given function \( g(z) \), if it can be written as \( (z-a)^{-n} h(z) \), where \( h(z) \) is analytic and \( h(a) not= 0 \), then \( z=a \) is a pole of order \( n \).In more practical terms, the order of a pole is just the number of times the function shoots towards infinity divided by that power of \(z\). From our exercise, for function \( f(z) = \frac{1 - \cosh z}{z^4} \), we expanded \( 1 - \cosh z \) to deduce it has a factor of \( -\frac{z^2}{2} \), leading us to conclude that the pole at \( z = 0 \) is of order 2. The Taylor series helped us see which terms were dominant near the pole and thus determine the order.

Hyperbolic Functions

Hyperbolic functions, including \( \sinh z \) and \( \cosh z \), are analogs of trigonometric functions, but they are based on hyperbolas instead of circles. They have various applications in many fields such as engineering and physics, and they can be defined using exponential functions. The basic hyperbolic function is:

• \( \cosh z = \frac{e^z + e^{-z}}{2} \)
• \( \sinh z = \frac{e^z - e^{-z}}{2} \)

These functions have unique properties, such as identities that mirror those of trigonometric functions, for example: \( \cosh^2 z - \sinh^2 z = 1 \). In our particular exercise, \( \cosh z \) was expanded around zero using Taylor series to help us observe its behavior near the origin. This inspection revealed which terms were important, leading to a recognition of the pole in the function \( f(z) \). Using these functions and their expansions often simplifies problems in complex analysis.