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. $$\frac{z+2}{z^{2}+9} \text { at } z=3 i$$

Short answer

The residue at \(z=3i\) is \(\frac{1 - 2i}{2}\).

Step-by-step solution

Identify the type of singularity

First, recognize that the point of interest is at \(z = 3i\). Determine whether this point is a simple pole or a higher-order pole for the function \(f(z) = \frac{z+2}{z^2 + 9}\).

Rewrite denominator

Notice that \(z^2 + 9 = (z - 3i)(z + 3i)\). Hence, the function can be rewritten as \(f(z) = \frac{z+2}{(z - 3i)(z + 3i)}\), identifying poles at \(z = 3i\) and \(z = -3i\).

Find the residue at the simple pole

Given that \(z = 3i\) is a simple pole, the residue can be found using the formula for the residue at a simple pole: \[ \text{Res}(f, 3i) = \frac{g(z)}{h'(z)} \bigg|_{z=3i} \] where \(g(z) = z + 2\) and \(h(z) = z^2 + 9\).

Calculate derivatives and evaluate

Find the derivative of the denominator function: \[ h'(z) = \frac{d}{dz}(z^2 + 9) = 2z \] Then evaluate this at \(z=3i\): \[ h'(3i) = 2(3i) = 6i \] Next, evaluate the numerator function at \(z=3i\): \[ g(3i) = 3i + 2 \]

Compute the residue

Putting it all together, compute the residue: \[ \text{Res}(f, 3i) = \frac{3i + 2}{6i} \] Simplify the fraction: \[ \text{Res}(f, 3i) = \frac{3i + 2}{6i} = \frac{3i + 2}{6i} \frac{i}{i} = \frac{(3i + 2)i}{6i^2} = \frac{-3 + 2i}{-6} = \frac{3 - 2i}{6} = \frac{1 - 2i}{2} \]

Key concepts

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex numbers. These functions are often represented as complex functions, which can include real and imaginary components. One of the primary topics in complex analysis is the behavior of functions within the complex plane, especially their continuity, differentiability, and integration.
To understand complex analysis, here are a few critical points to keep in mind:

• A complex number is generally expressed as: \(z = x + iy\), where \(x\) and \(y\) are real numbers, and \(i\) is the imaginary unit, satisfying \(i^2 = -1\).
• A complex function is a function that maps complex numbers to other complex numbers: \(f(z) = u(x, y) + iv(x, y)\), where \(u(x, y)\) and \(v(x, y)\) are real-valued functions.
• Key properties include differentiation (holomorphic functions) which requires satisfying the Cauchy-Riemann equations, and integration along complex paths, which forms the basis for residue calculus.

Mastery of these concepts allows deeper exploration into more advanced elements such as singularities and residue calculus.

Residue Theorem

The residue theorem is a powerful tool in complex analysis that simplifies the evaluation of complex integrals. It relates the integral of a function around a closed curve to the sum of residues of its singularities inside the curve.
Here's a simplified explanation:

• The residue at a point \( z_0 \) is a measure of the behavior of a function near that point. For functions with simple poles, this involves calculating the limit: \text{Res}(f,z_0) = \lim_{z \to z_0} (z - z_0)f(z).
• The theorem states that for a meromorphic function \( f \) and a positively oriented simple closed contour enclosing points \( z_1, z_2, \ldots, z_n \), the integral around the contour is equal to \(2\pi i\) times the sum of the residues: \oint_{C} f(z) dz = 2 \pi i \sum \text{Res}(f, z_i).
• Using the residue theorem simplifies complex integrals, especially when direct integration methods are cumbersome or infeasible.

This theorem is particularly used when evaluating integrals of functions with isolated singularities, which leads us to the next important topic: singularities.

Singularities

In complex analysis, singularities are points where a complex function fails to be analytic. Understanding singularities is crucial for analyzing and simplifying complex functions.
There are several types of singularities:

• Removable Singularities: Points where a function is not defined, but can be made analytic by appropriately defining or redefining the function at that point. For example, \(f(z) = \frac{\sin(z)}{z}\) has a removable singularity at \(z=0\).
• Poles: Points where a function's magnitude becomes unbounded. A function \(f(z)\) has a pole of order \(m\) at \(z_0\) if \(\lim_{z \to z_0} (z - z_0)^m f(z)\) is finite and non-zero. For example, \(f(z) = \frac{1}{(z - z_0)^n}\) has a pole of order \(n\) at \(z = z_0\).
• Essential Singularities: Points where a function exhibits highly erratic behavior. Laurent series near essential singularities include infinitely many negative power terms. For example, \(f(z) = e^{1/z}\) has an essential singularity at \(z = 0\).

In the given exercise, the function \(f(z) = \frac{z+2}{z^{2}+9}\) has isolated singularities (simple poles) at \(z=3i\) and \(z=-3i\). Analyzing these points and applying residue calculus allows us to compute integrals and solve complex problems effectively.