Question

. For the following, determine if the given point is a removable singularity, an essential singularity, or a pole (indicate its order). a. \(\frac{1-\cos z}{z^{2}}, \quad z=0\) b. \(\frac{\sin 2}{z^{2}}, \quad z=0\) c. \(\frac{z^{2}-1}{(z-1)^{2}}, \quad z=1\). d. \(z e^{1 / z}, \quad z=0\). e. \(\cos \frac{\pi}{\pi-\pi}, \quad z=\pi\)

Short answer

The point z=0 for the function \( \frac{1-\cos z}{z^{2}} \) is a removable singularity or a pole, determined by applying l'Hopital's rule. The point z=0 for the function \( \frac{\sin 2}{z^{2}} \) is a pole of order 2. The point z=1 for the function \( \frac{z^{2}-1}{(z-1)^{2}} \) is a removable singularity. The point z=0 for the function \( z e^{1 / z} \) is an essential singularity. And the point z=\( \pi \) for the function \( \cos \frac{\pi}{\pi-\pi} \) is an essential singularity.

Step-by-step solution

Identifying singularity for \( \frac{1-\cos z}{z^{2}}, \quad z=0\)

For this expression, apply l'Hopital's rule twice to find the limit as z approaches 0. The result of l'Hopital's rule for the function, if it exists, would indicate that the point z=0 is a removable singularity. If it does not exist, it would indicate that z=0 is a pole or an essential singularity.

Identifying singularity for \( \frac{\sin 2}{z^{2}}, \quad z=0\)

This function is not dependent on z, therefore the singularity at z = 0 is a pole of order 2.

Identifying singularity for \( \frac{z^{2}-1}{(z-1)^{2}}, \quad z=1\)

This function simplifies to \( \frac{(z-1)(z+1)}{(z-1)^{2}} \). After cancelling out \( z-1 \) from the numerator and the denominator, we get \( \frac{z+1}{z-1} \) which is defined at z = 1. This indicates that point z = 1 of the original function is a removable singularity.

Identifying singularity for \( z e^{1 / z}, \quad z=0\)

This function can't be simplified using algebraic manipulations, nor can use l'Hopital's rule to find a limit as z approaches 0. This indicates that the singularity at z = 0 is an essential singularity.

Identifying singularity for \( \cos \frac{\pi}{\pi-\pi}, \quad z=\pi\)

This function simplifies to \( \cos (1/0) \), which is undefined for any real number. Therefore, the singularity at z = \( \pi \) is an essential singularity.

Key concepts

Removable Singularity

A removable singularity is a point on the complex plane where a function is not defined, but the limit of the function as it approaches that point exists and is finite. Conceptually, you can 'remove' the singularity just by defining the function at that point to be equal to this finite limit.

In the context of our exercise, a removable singularity is observed when, after certain algebraic manipulations or cancellation, the function becomes well-defined at that point. For example, the function \(\frac{z^2 - 1}{(z - 1)^2}\), becomes \(\frac{z + 1}{z - 1}\) after reducing, thus clarifying that \(z = 1\) is not actually problematic for the function, and hence it's a removable singularity.

Pole of Function

A pole of a function is a type of singularity that causes the function's value to approach infinity as the variable approaches the pole's location. The order of the pole is determined by how quickly the function approaches infinity; technically, it's the power of \(z - z_0\) in the denominator that causes the blowup at the pole \(z_0\).

For instance, for \(\frac{\sin 2}{z^2}\), as \(z\) approaches 0, the function skyrockets to infinity, characterizing \(z = 0\) as a pole of order 2. It is crucial to distinguish a pole from other singularities as it signifies not just a missing point but a 'tall spike' in the complex plane.

Essential Singularity

An essential singularity is more enigmatic than a removable singularity or a pole. At an essential singularity, the behavior of the function is vastly unpredictable, and the limits can oscillate wildly or not exist at all. According to the complex version of Picard's Theorem, near an essential singularity, a complex function takes on every possible complex value, with possibly one exception, infinitely often.

The function \(z e^{1/z}\) from our exercises has an essential singularity at \(z = 0\). The absence of a well-defined limit and the impossibility of a cancellation mark the singularity as essential. Similarly, the function \(\cos (1/0)\) yields an undefined value, exemplifying the erratic nature of essential singularities.

L'Hopital's Rule

L'Hopital's Rule is a magnificent tool when dealing with indeterminate forms like 0/0 or \(\infty/\infty\). The rule states if you have a limit that yields an indeterminate form, you can take the derivative of the numerator and the derivative of the denominator and then reevaluate the limit.

Applying L'Hopital's Rule for the function \(\frac{1 - \cos z}{z^2}\) at \(z=0\), we can differentiate the numerator and denominator respectively. If this new limit exists and is finite, our original singularity is removable. This is precisely what occurs in step one of the exercise, illustrating how L'Hopital's Rule aids in classifying singularities.