Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{e^{12}}{\left(z^{2}+4\right)^{2}} \text { at } z=2 i $$

Short answer

-\frac{e^{12}}{32}i

Step-by-step solution

Identify the singularity

The given function is \(\frac{e^{12}}{(z^2+4)^2}\) at \(z = 2i\). First, rewrite it as \(\frac{e^{12}}{[(z - 2i)(z + 2i)]^2}\). The singularity is at \(z = 2i\).

Determine the order of the pole

The denominator \((z^2 + 4)^2\) or \((z - 2i)^2(z + 2i)^2\) indicates that \(z = 2i\) is a pole of order 2.

Calculate the residue using the second-order pole formula

For a second-order pole, the residue is calculated as: \[\text{Res}(f, z_0) = \frac{1}{1!} \lim_{{z \to z_0}} \frac{d}{dz} \left[(z-z_0)^2 f(z)\right] \]In this case, \[(z - 2i)^2 \frac{e^{12}}{(z^2 + 4)^2} = \frac{e^{12}}{(z + 2i)^2}\].

Differentiate the simplified term

Differentiate \(\frac{e^{12}}{(z + 2i)^2}\) with respect to \(z\). \(\frac{d}{dz} \frac{e^{12}}{(z + 2i)^2} = e^{12} \frac{d}{dz} (z + 2i)^{-2} = e^{12} (-2)(z + 2i)^{-3} = -\frac{2e^{12}}{(z + 2i)^3}\).

Substitute the point \(z = 2i\)

Substitute \(z = 2i\) back into the differentiated term: \[-\frac{2e^{12}}{(2i + 2i)^3} = -\frac{2e^{12}}{(4i)^3} = -\frac{2e^{12}}{64i^3} = -\frac{e^{12}}{32i^3}\]Since \(i^3 = -i\), this simplifies to: \[-\frac{e^{12}}{32(-i)} = \frac{e^{12}}{32i}\].

Simplify the expression

To further simplify, multiply by \(i\) in the numerator and denominator: \[\frac{e^{12}i}{32i^2} = \frac{e^{12}i}{32(-1)} = -\frac{e^{12}i}{32} = -\frac{e^{12}}{32}i\].

Key concepts

Residue Calculus

Residue calculus is a powerful tool in complex analysis. It helps us solve problems, such as computing complex integrals or determining residues at singular points. The residue of a function at a point is the coefficient of \(\frac{1}{z-z_0}\) in its Laurent series expansion around that point.

Here's how to apply residue calculus:

• Identify singularities: Locate points where the function is not analytic (i.e., points at which the function could be undefined).
• Determine the order of the pole: Decide whether it's a simple pole (order 1), or a higher-order pole (order 2 or more).
• Use the residue formula based on the order of the pole to calculate the residue. For a second-order pole, the residue formula involves differentiation.

Let's delve deeper into each of these steps with specific concepts.

Poles and Singularities

In complex analysis, singularities are points where a complex function is not defined or not analytic. These can be classified into:

• Poles: A pole of order \(n\) at \(z_0\) means the function behaves like \(\frac{1}{(z-z_0)^n}\) near \(z_0\).
• Essential Singularities: More complicated and do not resemble poles, where the function displays erratic behavior.
• Removable Singularities: Points where the function can be redefined to render it analytic.

In our example, the function \(\frac{e^{12}}{(z^2+4)^2}\) has a pole of order 2 at \(z=2i\). A higher-order pole means more complex residue calculations.

Understanding the type and order of singularities helps in choosing the right technique for computing residues.

Differentiation in Complex Plane

Differentiation in the complex plane works similarly to real differentiation but requires understanding of complex variables. For a second-order pole, to find residues, we take the derivative.
For instance, given the function \(\frac{e^{12}}{(z+2i)^2}\), we simplify when finding the residue at \(z=2i\). First, identify \(f(z)\), then differentiate: \( \frac{d}{dz} \frac{e^{12}}{(z+2i)^2} = e^{12} \frac{d}{dz} (z+2i)^{-2} = e^{12} (-2)(z+2i)^{-3} \).

Differentiation is essential here as it simplifies the calculation of residues for higher-order poles. Continuing the computation involves substituting \(z=2i\) into the derived expression and simplifying.

Second-Order Poles

Understanding second-order poles is crucial. A second-order pole means the function has a pole where the term's exponent in the denominator is squared. The residue at a second-order pole requires a differentiated function multiplied by \( (z-z_0)^2 \): \[ \text{Res}(f, z_0) = \frac{1}{1!} \frac{d}{dz} \big[(z-z_0)^2 f(z)\big] \bigg|_{z=z_0} \]

Using our example function:

• Rewrite it as needed, for example: \(\frac{e^{12}}{[(z-2i)(z+2i)]^2}\).
• Differentiate: We derived \( e^{12} (-2)(z+2i)^{-3} \).
• Substitute \(z=2i\) and simplify: The residue found was \( -\frac{e^{12}i}{32}\).

Each step aids in resolving residues for complex functions with higher-order poles.