Problem 4 - A First Course in Complex Analysis with Applications

Question

Show that \(z=0\) is a removable singularity of the given function. Supply a definition of \(f(0)\) so that \(f\) is analytic at \(z=0\). $$ f(z)=\frac{1-\frac{1}{2} z^{10}-\cos z^{5}}{\sin z^{2}} $$

Step-by-Step Solution

Answer

Define \(f(0) = 0\) to remove the singularity and make \(f\) analytic at \(z=0\).

Step 1: Identify the Singularity

The function \(f(z)=\frac{1-\frac{1}{2} z^{10}-\cos z^{5}}{\sin z^{2}}\) has a singularity at \(z=0\) because the denominator \(\sin z^2\) becomes zero when \(z=0\). This indicates a potential singularity that may be removable depending on the behavior of the numerator.

Step 2: Expand Numerator in Taylor Series

The numerator is a combination of a polynomial and a cosine function. The Taylor series expansion for \(\cos z^5\) around \(z=0\) is \(1 - \frac{(z^5)^2}{2!} + \cdots = 1 - \frac{z^{10}}{2} + \cdots\). Substituting this into the numerator, we get:\[ 1 - \frac{1}{2}z^{10} - \left(1 - \frac{z^{10}}{2} + \cdots \right). \]Upon simplification, this results in the numerator \(z^{10}\).

Step 3: Expand Denominator in Taylor Series

The Taylor series expansion of \(\sin z^2\) around \(z=0\) is \(z^2 - \frac{(z^2)^3}{3!} + \cdots = z^2 - \frac{z^6}{6} + \cdots \). This means that as \(z\) approaches zero, \(\sin z^2\) behaves approximately like \(z^2\).

Step 4: Cancel Out Terms

The function becomes \[\frac{z^{10}}{z^2 - \frac{z^6}{6} + \cdots}.\]To simplify this expression, observe that the leading order term in the denominator is \(z^2\). Thus, \[\frac{z^{10}}{z^2} = z^8 .\]

Step 5: Determine Analytic Extension

Since \(f(z)\) reduces to \(z^8\) when \(z eq 0\), we can define \(f(0) = 0\) to ensure that \(f(z)\) is continuous and therefore analytic at \(z=0\). Here \(z=0\) becomes a removable singularity as \(f(z) = z^8\) is analytic everywhere.

Key Concepts

Removable Singularity

A removable singularity occurs when a function is not defined at a point, but can be defined there such that the function becomes analytic. For a complex function, this means that while the function may initially appear to be undefined at a point due to a zero in the denominator, we can "remove" this singularity by redefining the function's value at that point.

In the exercise, the point 0 for the function \( f(z) = \frac{1-\frac{1}{2} z^{10}-\cos z^{5}}{\sin z^{2}} \) is initially a candidate for a singularity because the denominator, \( \sin z^2 \), is zero at \( z=0 \). However, if the numerator also goes to 0, the singularity might be removable. By simplifying the expression, we found that near \( z=0 \), both the numerator and the denominator contain powers of \( z \) that allow these terms to cancel each other out.

This results in redefining the function at \( z=0 \), such that it becomes whole and analytic. Specifically, the function can be defined as \( f(0)=0 \), eliminating the indeterminate form and making the singularity removable.

Taylor Series Expansion

The Taylor series is a way to represent a function as an infinite sum of terms based on its derivatives at a single point. This is often used to approximate functions and analyze their behavior near that point. In complex analysis, Taylor series expansions are especially useful around singular points.

In the exercise, the Taylor series played a crucial role in simplifying the numerator and denominator of the function. The numerator \( 1 - \frac{1}{2} z^{10} - \cos z^{5} \) was expanded using the Taylor series for \( \cos z^5 \):

• The series for \( \cos z^5 \) is \( 1 - \frac{(z^5)^2}{2!} + \cdots \).
• Substituting this into the original expression gives \( 1 - \frac{z^{10}}{2} \), leading to the cancellation of terms.

Similarly, the denominator \( \sin z^2 \) was expanded:

• The series for \( \sin z^2 \) is \( z^2 - \frac{(z^2)^3}{3!} + \cdots \).

These substitutions and simplifications reveal how powers of \( z \) interact, facilitating the removal of the singularity and finding the simplest form of the function.

Analytic Functions

An analytic function is a function that is locally given by a convergent power series. This means that around any point in its domain, the function can be written as a power series. Such functions are infinitely differentiable and have derivatives of all orders.

In this exercise, showing that the function is analytic at \( z=0 \) meant demonstrating that the function's behavior can be described seamlessly by a power series.
The function \( f(z) \), after using Taylor series expansions and removing the singularity, is seen to be \( z^8 \) at and near \( z=0 \).

The simplicity of \( z^8 \) shows that it's not only continuous but also differentiable at all orders, thereby confirming its analyticity at \( z=0 \). Additionally, by defining \( f(0) = 0 \), we ensure continuity and provide a smooth, analytic extension over the entire domain of interest.