Problem 28 - A First Course in Complex Analysis with Applications

Question

Show that the indicated number is an essential singularity of the given function. $$ f(z)=(z-1) \cos \left(\frac{1}{z+2}\right) ; z=-2 $$

Step-by-Step Solution

Answer

The function has an essential singularity at \( z = -2 \) due to the infinite negative powers in its series expansion.

Step 1: Identify Singularity Type

An essential singularity occurs when a function's Laurent series has infinitely many negative power terms. At \( z = -2 \), the term \( \frac{1}{z+2} \) becomes undefined, indicating a singularity there. We will verify if it is essential by analyzing the series expansion.

Step 2: Series Expansion of Core Function

Consider the function \( \cos \left( \frac{1}{z+2} \right) \). The expansion of \( \cos(x) \) is given by \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \). Substituting \( x = \frac{1}{z+2} \), we have:\[ \cos \left( \frac{1}{z+2} \right) = 1 - \frac{1}{2} \left( \frac{1}{z+2} \right)^2 + \frac{1}{24} \left( \frac{1}{z+2} \right)^4 - \cdots \]This series has infinitely many negative powers of \( z+2 \).

Step 3: Analyze the Full Function

Combine the series expansion with \( (z-1) \). We have:\[f(z) = (z-1) \left[ 1 - \frac{1}{2(z+2)^2} + \frac{1}{24(z+2)^4} - \cdots \right]\]When multiplied out, each term \( (z-1) \left( \frac{1}{(z+2)^n} \right) \) will still have negative powers coming from the \( \cos \) expansion, confirming it does not cease after finitely many terms.

Step 4: Conclusion on Singularity

Since the series expansion of \( f(z) \) around \( z = -2 \) has infinitely many negative powers, the singularity at \( z = -2 \) is indeed essential. The behavior of the function \( f(z) \) near \( z = -2 \) is highly erratic, typical of an essential singularity.

Key Concepts

Laurent Series

The Laurent series is a powerful tool used in complex analysis, particularly for dealing with functions that have singular points. Unlike the Taylor series, which only contains non-negative integer power terms, the Laurent series can include terms that are negative powers of variables.
These negative terms allow the Laurent series to represent functions near singularities. In the context of complex numbers, if you have a function with a singularity, it means the function behaves differently in a neighborhood of that point. The Laurent series expands the function into a series of powers around such points.

• The general form is: \[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z-c)^n \]
• The Laurent series can show whether a singularity is removable, a pole, or an essential singularity.
• An essential singularity has infinitely many negative powers, like seen in the given function example: \( f(z) = (z-1) \cos \left(\frac{1}{z+2}\right) \).

Complex Functions

Complex functions use complex variables, forming the foundation for many advanced fields such as quantum physics and electrical engineering. A function is said to be complex if its input and output are both complex numbers.
These functions can exhibit rich and interesting behavior, unlike their real counterparts. For example, they can possess singularities, which are points where the function does not behave normally.

• One common type of singularity is an essential singularity, where the behavior of the function becomes highly unpredictable.
• At these points, a function can map small neighborhoods around the singularity to almost the entire complex plane, as significantly and sharply demonstrated by Picard's theorem.
• In the example \( f(z) = (z-1) \cos \left(\frac{1}{z+2}\right) \), the function has an essential singularity at \( z=-2 \).

Series Expansion

Series expansion is a method of expressing a function as a sum of terms, calculated from the values of its derivatives at a single point.
In complex analysis, this technique helps in approximating behaviors of functions around specified points, especially around singularities.

• When a function such as \( \cos(x) \) is expanded, it is represented by a series: \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]
• Substituting \( x = \frac{1}{z+2} \) shows how \( \cos \left( \frac{1}{z+2} \right) \) results in a Laurent series containing negative powers \((\frac{1}{(z+2)^2})\), confirming complex behavior near \( z=-2 \).
• The expanded series helps identify the function's essential singularity by showcasing these infinite negative powers.