Problem 24 - A First Course in Complex Analysis with Applications

Question

Determine the order of the poles for the given function. $$ f(z)=\frac{e^{z}-1}{z^{2}} $$

Step-by-Step Solution

Answer

The function has a simple pole of order 1 at \( z = 0 \).

Step 1: Understand the Expression

We need to determine the order of the poles for the function \( f(z) = \frac{e^z - 1}{z^2} \). A pole is a point at which a function goes to infinity.

Step 2: Identify the Problematic Point

Examine the denominator, \( z^2 \). There is a potential pole at \( z = 0 \) because the denominator becomes zero, making the function undefined.

Step 3: Analyze the Numerator

The numerator is \( e^z - 1 \). The Taylor series expansion for \( e^z \) around \( z = 0 \) is \( e^z = 1 + z + \frac{z^2}{2} + \cdots \). Thus, \( e^z - 1 = z + \frac{z^2}{2} + \cdots \), which simplifies to \( z(1 + \frac{z}{2} + \cdots) \).

Step 4: Determine the Order of the Pole

Rewrite the function using the expansion obtained. Substitute the expansion of the numerator into the function: \( f(z) = \frac{z(1 + \frac{z}{2} + \cdots)}{z^2} = \frac{1 + \frac{z}{2} + \cdots}{z} \). We see that the smallest power of \( z \) in the denominator is larger than the smallest power of \( z \) in the numerator, confirming the presence of a pole at \( z = 0 \).

Step 5: Compute the Order

The restructured function \( \frac{1 + \frac{z}{2} + \cdots}{z} \) shows that as \( z \to 0 \), the term \( \frac{1}{z} \) dominates. Since this term is the only one causing \( f(z) \to \infty \) at \( z = 0 \), the order of the pole is \( 1 \).

Key Concepts

Poles of a function

When studying complex functions, understanding the concept of poles is essential. A pole is a specific type of singularity where a function's value approaches infinity. Consider a function like \( f(z) \). If at a point \( z = z_0 \) there is a pole, then the function becomes undefined because the denominator of a fraction tends to zero.

To locate poles, follow these steps:

• Identify the points where the denominator equals zero.
• Analyze the numerator to see if these points cause indeterminate forms like \( \frac{0}{0} \).

This distinction helps determine if you have a pole or if the numerator cancels the effect, leading to a removable discontinuity instead.

For example, consider the function \( f(z) = \frac{e^z - 1}{z^2} \). We see a potential problem point at \( z = 0 \) because plugging zero into \( z^2 \) in the denominator yields an undefined expression. This indicates a pole at \( z = 0 \).

Order of poles

Not only is it crucial to identify a pole, but understanding its order deepens our analysis. The order of a pole gives us information about how rapidly a function approaches infinity near that point.

To find the order of a pole:

• Conduct a factor analysis of the numerator and denominator around the pole.
• The order is the difference in the power of \( z \) in the denominator minus the power in the numerator.

Referring back to \( f(z) = \frac{e^z - 1}{z^2} \), note after simplifying the numerator’s Taylor series \( e^z - 1 = z + \frac{z^2}{2} + \cdots \), it helps express the function as \( \frac{z(1 + \frac{z}{2} + \cdots)}{z^2} \) which simplifies to \( \frac{1}{z} (1 + \frac{z}{2} + \cdots) \).

The smallest power of \( z \) here is in the denominator. Thus, this function has a dominant \( \frac{1}{z} \) term, confirming that the pole at \( z = 0 \) is of order 1.

Taylor series expansion

The Taylor series is a powerful tool in complex analysis that allows us to approximate functions near a given point. It expresses a function as an infinite sum of terms calculated from the function's derivatives at a certain point.

The Taylor series expansion around \( z = 0 \) (also called a Maclaurin series) for the exponential function \( e^z \) is:\[e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\]This series allows us to approximate \( e^z - 1 \) as \( z + \frac{z^2}{2} + \cdots \).

This expansion is crucial when evaluating expressions like \( \frac{e^z - 1}{z^2} \). When simplifying, it helps to isolate the lowest power of \( z \) in the numerator, ensuring that calculations of poles and their orders are accurate. Thus, Taylor expansions provide clarity by rewriting functions in a manageable form that identifies both the effects of singularities and convergence behavior near those points.