Problem 2 - Advanced Engineering Mathematics

Question

In Problems 1 and 2 , show that \(z=0\) is a removable singularity of the given function. Supply a definition of \(f(0)\) so that \(f\) is analytic at \(z=0\). $$ f(z)=\frac{\sin 4 z-4 z}{z^{2}} $$

Step-by-Step Solution

Answer

The singularity at \( z=0 \) is removable; define \( f(0) = 0 \) to make \( f \) analytical at \( z=0 \).

Step 1: Understand the given function

We have the function \( f(z) = \frac{\sin 4z - 4z}{z^2} \). This function will not be analytic at \( z = 0 \) due to division by zero. We aim to determine if this singularity is removable.

Step 2: Use power series expansion

To analyze the singularity at \( z = 0 \), we'll expand \( \sin(4z) \) into its power series: \( \sin(4z) = 4z - \frac{(4z)^3}{3!} + \frac{(4z)^5}{5!} - \cdots \). Subtracting \( 4z \) from this series gives us \( \sin(4z) - 4z = -\frac{(4z)^3}{3!} + \frac{(4z)^5}{5!} - \cdots \).

Step 3: Simplify using Taylor series

Substitute the series expansion from Step 2 back into the original function: \( f(z) = \frac{-\frac{(4z)^3}{3!} + \frac{(4z)^5}{5!} - \cdots}{z^2} \). Divide each term in the series by \( z^2 \): \( f(z) = -\frac{64z^2}{3!} + \frac{(4z)^3}{5!z^2} - \cdots = -\frac{64z}{6} + \frac{1024z^3}{120z^2} - \cdots \). Simplifying further, we get \( f(z) = -\frac{64}{6}z + \cdots \).

Step 4: Determine if singularity is removable

Notice that the lowest power term is \( -\frac{64}{6}z \), indicating that terms with \( z^2 \) and higher vanish when \( z = 0 \). Therefore, \( z=0 \) itself doesn't cause a problem in terms of convergence if appropriately defined.

Step 5: Define the function at z=0

To make \( f \) analytic at \( z=0 \), ensure that \( f(0) \) is equal to the limit as \( z \) approaches 0: \( f(0) = \lim_{z \to 0} \frac{\sin 4z - 4z}{z^2} = 0 \), thus making the function smooth and removing the singularity.

Key Concepts

Removable Singularities

A singularity in complex analysis is a point where a function is not defined or not differentiable. A removable singularity is a specific type where, despite the initial undefined state at that point, the singularity can be "removed" by redefining the function appropriately. For instance, consider the function given in the problem:

• The function is originally undefined at \( z = 0 \) due to division by zero.
• Through mathematical manipulation such as power series expansion or limits, one can redefine the value at this point.
• This allows the function to become smooth and analytic, meaning differentiable everywhere, including at the singularity's location.

In our context, redefining \( f(0) \) with the limit \( \lim_{z \to 0} \frac{\sin 4z - 4z}{z^2} = 0 \) ensured that the singularity is removable, and the function becomes analytic across the entire complex plane.

Analytic Functions

An analytic function in complex analysis is a function that is locally given by a convergent power series. This means the function can be expressed as a power series around any point within its domain, where the coefficients are derived from the function's derivatives at that point. Analyticity is a strong form of differentiability that implies:

• The function is differentiable not just once, but infinitely many times.
• The power series expansion matches the function exactly within a region of convergence.

An example is the function \( f(z) = e^z \), which can be expanded into its power series everywhere it is defined. For our function \( f(z) = \frac{\sin 4z - 4z}{z^2} \), after removing the singularity at \( z = 0 \), it is analytic since it can be represented by a power series in the neighborhood of that point.

Power Series Expansion

A power series is an infinite series of the form: \[S(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + \cdots\]where each \( a_n \) is a coefficient and \( z \) is the variable. In complex analysis, power series are crucial since they can represent functions around points where they are analytic. The key elements include:

• The range of \( z \) where the series converges is called the radius of convergence.
• For every point within this radius, the series provides an accurate representation of the function.
• In the problem, we expanded \( \sin(4z) \) into a power series, which helped identify the removable singularity at \( z=0 \).

By expanding \( \sin(4z) = 4z - \frac{(4z)^3}{3!} + \cdots \) minus \( 4z \), we simplified the problematic term and facilitated finding a smooth extension of \( f(z) \).

Taylor Series

The Taylor series is a special classification of power series and is used to approximate functions near a particular point using derivatives. It is written as:\[T(z) = f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \frac{f'''(a)}{3!}(z-a)^3 + \cdots\]where \( a \) is the center of the series, and \( f^n(a) \) are the derivatives of the function evaluated at \( a \). The attributes of a Taylor Series include:

• Convergence around the point \( a \) where the function is analytic.
• Exact representation of the function within its radius of convergence.
• It can also be extended further to become a Maclaurin series when \( a = 0 \).

In the given exercise, expanding \( \sin(4z) \) into a Taylor series allowed us to analyze the terms that contribute to the singularity and define \( f(0) \) smoothly, consequently removing the singularity at \( z=0 \).