Question

For each of the following functions, say whether the indicated point is regular, an essential? singularity, or a pole, and if a pole of what order it is. (a) \(\frac{e^{z}-1-z}{z^{2}}, \quad z=0\) (b) \(\frac{\sin z}{z^{3}}, \quad z=0\) (c) \(\frac{z^{2}-1}{(z-1)^{2}}, z=1\) (d) \(\frac{\cos z}{(z-\pi / 2)^{4}}, z=\pi / 2\)

Short answer

(a) Regular at \(z=0\). (b) Pole of order 2 at \(z=0\). (c) Pole of order 1 at \(z=1\). (d) Pole of order 4 at \(z=\frac{\pi}{2}\).

Step-by-step solution

Part (a) - Identify the type of singularity for \( \frac{e^{z} - 1 - z}{z^{2}}, \quad z=0 \)

Write \( f(z) = \frac{e^{z} - 1 - z}{z^2} \). Use the Maclaurin series for \( e^z \), which gives \( e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \). Subtract \(1\) and \(z\) from this series to get: \( e^z - 1 - z = \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \). Then, \( f(z) = \frac{\frac{z^2}{2!} + \frac{z^3}{3!} + \cdots}{z^2} = \frac{1}{2!} + \frac{z}{3!} + \cdots \). This expression is regular at \(z=0\).

Part (b) - Identify the singularity for \( \frac{\sin z}{z^{3}}, \quad z=0 \)

Write \( f(z) = \frac{\sin z}{z^3} \). Use the Maclaurin series for \( \sin z \), which gives \( \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots \). Substitute this into the expression to get: \( f(z) = \frac{z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots}{z^3} = \frac{1}{z^2} - \frac{z}{3!} + \cdots \). This indicates a pole of order 2 at \(z=0\).

Part (c) - Identify the singularity for \( \frac{z^2 - 1}{(z - 1)^2}, z=1 \)

Write \( f(z) = \frac{z^2 - 1}{(z-1)^2} \). Factor the numerator: \( z^2 - 1 = (z-1)(z+1) \). So, \( f(z) = \frac{(z-1)(z+1)}{(z-1)^2} = \frac{z+1}{z-1} \). As \( z \rightarrow 1 \), this simplifies to \( f(z) = \frac{2}{z-1} \), indicating a pole of order 1 at \(z=1\).

Part (d) - Identify the singularity for \( \frac{\cos z}{(z-\frac{\pi}{2})^{4}}, z=\frac{\pi}{2} \)

Write \( f(z) = \frac{\cos z}{(z - \frac{\pi}{2})^4} \). Use the Maclaurin series for \( \cos z \). At \( z = \frac{\pi}{2} \), \( \cos(z) = 0 \) and \( \cos'(z) = -\sin(z) \cdots \). Use series expansion methods if necessary. Since \( \cos(\frac{\pi}{2}) = 0 \), we get that \( \frac{\cos z}{(z - \frac{\pi}{2})^4} \) implies a regular function divided by \( (z - \frac{\pi}{2})^4 \), confirming a pole of order 4 at \( z = \frac{\pi}{2} \).

Key concepts

regular point

The term 'regular point' in complex analysis refers to a point where the function is analytic (differentiable) and doesn't have any singularities. Simply put, at a regular point, a function behaves nicely and can be expressed as a convergent power series.
For example, take the function \( \frac{e^z - 1 - z}{z^2} \) at \( z = 0 \). By expanding \( e^z \) using its Maclaurin series:
\[ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \text{...} \] and then subtracting 1 and \( z \), we have:
\[ e^z - 1 - z = \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \text{...} \] Dividing by \( z^2 \) to get \( f(z) \), we obtain:
\[ f(z) = \frac{\frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \text{...}}{z^2} = \frac{1}{2!} + \frac{z}{3!} + \frac{z^2}{4!} + \text{...} \]
Since this last expression still converges at \( z = 0 \) and does not exhibit a singularity, \( z = 0 \) is a regular point for this function.

essential singularity

An essential singularity is a point where the function's behavior is extremely unpredictable. Around this point, the values of the function can become very erratic. Mathematically, Laurent series about such points have infinitely many non-zero terms with negative powers.
For instance, if the function had an essential singularity at \( z = 0 \), the behavior would not be captured by a simple series as examined in regular points or poles. Instead, given any complex number, there is some sequence approaching the singularity that makes the function take on that value. However, in the provided exercise examples, none of the functions have an essential singularity. They either have a regular point or a pole.

pole

A pole of a function happens when the function heads towards infinity as it approaches a specific point. Poles are classified based on their order, equivalent to how fast the function grows near the pole. A simple pole (order 1) means the function grows like \( \frac{1}{z} \) near the point, while higher-order poles grow faster.
Let's look at an example from the exercise:
For \( \frac{\text{sin} z}{z^3}, \text{ at } z=0 \), write \( f(z) = \frac{\text{sin} z}{z^3} \). Using the Maclaurin series for \(\text{sin} z \), we get:
\[ \text{sin} z = z - \frac{z^3}{3!} + \frac{z^5}{5!} + \text{...} \] Thus,
\[ f(z) = \frac{z - \frac{z^3}{3!} + \frac{z^5}{5!} - \text{...}}{z^3} = \frac{1}{z^2} - \frac{z}{3!} + \frac{z^3}{5!} - \text{...} \]
This shows a term \( \frac{1}{z^2} \) which makes \( z=0 \) a pole of order 2. Another example is \( \frac{\text{cos} z}{(z-\frac{\text{π}}{2})^4} \) at \( z = \frac{\text{π}}{2} \). Since \( \text{cos}(\frac{\text{π}}{2}) = 0 \), the Maclaurin series around \( z=\frac{\text{π}}{2} \) is zero, making it a pole of order 4 given the divisor \((z-\frac{\text{π}}{2})^4 \).