Question

Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer. $$\frac{e^{i z}}{\left(z^{2}+4\right)^{2}} \text { at } z=2 i$$

Short answer

\(\frac{e^{-2}}{8i}\).

Step-by-step solution

Identify the Type of Singularity

Determine the type of singularity for the function at the given point. The function \(\frac{e^{i z}}{\left(z^{2}+4\right)^{2}}\) at \(z=2i\) has a pole of order 2 since \(\left(z^{2}+4\right)^{2} = (z-2i)^2(z+2i)^2\) in the denominator.

Expand the Function Around the Point

Consider the Laurent series expansion around \(z=2i\). For a pole of order 2 at \(z=2i\), the residue of the function takes a specific form. Write \(z = 2i + w\), then substitute \(z = 2i + w\) into the function and expand.

Simplify the Expression

Substitute \(z = 2i + w\) into \(e^{iz}\) to get \(e^{i(2i+w)} = e^{-2}e^{iw}\). The denominator becomes \((w(4i+w)^2)\). The function simplifies to \frac{e^{-2} e^{iw}}{w^2 \times 2^2 \times (1 + \frac{w}{4i})^2}\.

Extract the Coefficient of \(\frac{1}{w}\)

Use the series expansion for \(\frac{1}{(1+\frac{w}{4i})^2}\) which expands to \1 - \frac{w}{2i} + o(w^2)\. Collect all \(\frac{1}{w}\) terms after substituting back.

Calculate the Residue

Combine all terms and extract the residue: \(\text{Res}(\frac{e^{iz}}{(z^2+4)^2}, 2i) = e^{-2} \frac{1}{8i} = \frac{e^{-2}}{8i}\).

Key concepts

complex analysis

Complex analysis is a branch of mathematics that studies functions of complex variables. These functions have complex numbers as their domain and range. Complex numbers are expressed as \(a + bi\), where \(i\) is the imaginary unit, and \(a\) and \(b\) are real numbers. This field deals with concepts such as limits, continuity, differentiation, and integration but in the complex plane.
When you study complex analysis, you encounter many powerful theorems and tools that are not available when dealing only with real numbers. These include:

• Analytic Continuation
• Cauchy's Integral Theorem
• Cauchy's Integral Formula
• Residue Theorem
• Laurent Series

The Residue Theorem is particularly useful in evaluating complex integrals and understanding the behavior of functions near their singularities.

Laurent series

The Laurent series is a representation of a complex function as a power series, which includes terms with negative exponents. It’s similar to the Taylor series but more general. This series allows a function to be expressed near a singularity. For a function \(f(z)\) around a point \(z_0\), the Laurent series looks like this:
\[ f(z) = \frac{a_{-n}}{(z-z_0)^n} + \frac{a_{-n+1}}{(z-z_0)^{n-1}} + \frac{a_0}{(z-z_0)^0} + a_1 (z-z_0) + a_2 (z-z_0)^2 + \text{...} \]
Here, \(a_{-n}\) are coefficients, and the term \(\frac{a_{-1}}{(z-z_0)}\) is particularly important as it relates to the concept of residue. The residue is the coefficient of \(\frac{1}{z - z_0}\) in the Laurent series, which plays a crucial role in evaluating complex integrals using the Residue Theorem.

singularity

A singularity is a point at which a complex function ceases to be analytic, meaning it cannot be expressed as a convergent power series. Types of singularities include:

• Removable Singularities
• Poles
• Essential Singularities

A point \(z = z_0\) where a function becomes undefined, or 'blows up,' is considered a singularity of the function. Identifying and understanding the type of singularity is important for analyzing a function's behavior near that point. The Laurent series helps to characterize these behaviors.
For example, in the problem provided, \(z = 2i\) is a pole of order 2 for the function \(\frac{e^{iz}}{(z^2 + 4)^2}\).

poles

Poles are a specific type of singularity where a function's magnitude tends to infinity as the variable approaches the pole. A function \(f(z)\) has a pole of order \(m\) at \(z = z_0\) if:
\[\lim_{z \to z_0} (z - z_0)^m f(z) \text{ is finite and nonzero}\]
Poles are classified by their order, the number of terms in the Laurent series needed for the function to 'blow up.'
In our example, at \(z = 2i\), the function \(\frac{e^{iz}}{(z^2 + 4)^2}\) has a pole of order 2, meaning when \((z - 2i)^2\) is multiplied, the function becomes finite. This insight simplifies finding the residue, which is crucial for evaluating integrals using the Residue Theorem. Identifying pole order helps in expanding the function in its Laurent series and calculating residues effectively.