Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{1}{(1-2 z)(5 z-4)} $$ at \(z=\frac{1}{2}\) and at \(z=\frac{4}{5}\)

Short answer

The residues are \(-\frac{2}{3}\) at \(z = \frac{1}{2}\) and \(-\frac{5}{3}\) at \(z = \frac{4}{5}\).

Step-by-step solution

Identify Singularities

Determine the points at which the function has singularities. The singularities are the points where the denominator of the function equals zero.

Solve for Singularities

Solve the equations from Step 1:\((1 - 2z) = 0\) gives \(z = \frac{1}{2}\) and \((5z - 4) = 0\) gives \(z = \frac{4}{5}\).

Determine the Type of Singularity

The singularities at \(z = \frac{1}{2}\) and \(z = \frac{4}{5}\) are simple poles because the denominators are first-order terms.

Find Residue at \(z = \frac{1}{2}\)

Use the residue formula for a simple pole: \( \text{Res}(f, a) = \lim_{z \to a} (z - a) f(z) \). Find the residue:\[\begin{align*}\text{Res}\left( \frac{1}{(1-2z)(5z-4)}, \frac{1}{2} \right) &= \lim_{z \to \frac{1}{2}} (z - \frac{1}{2}) \frac{1}{(1-2z)(5z-4)} \&= \lim_{z \to \frac{1}{2}} (z - \frac{1}{2}) \frac{1}{(1-2z)(5z-4)} \&= \lim_{z \to \frac{1}{2}} \frac{(z - \frac{1}{2})}{(1-2z)(5z-4)} \&= \lim_{z \to \frac{1}{2}} \frac{1}{5z-4} \cdot \lim_{z \to \frac{1}{2}} \frac{z - \frac{1}{2}}{1-2z} \&= \frac{1}{5(\frac{1}{2})-4} \cdot \lim_{z \to \frac{1}{2}} \frac{1}{-2} \&= \frac{1}{\frac{5}{2}-4} \cdot \frac{1}{-2} \&= -\frac{2}{3}.\end{align*}\]

Find Residue at \(z = \frac{4}{5}\)

Similarly, find the residue:\[\begin{align*}\text{Res}\left( \frac{1}{(1-2z)(5z-4)}, \frac{4}{5} \right) &= \lim_{z \to \frac{4}{5}} (z - \frac{4}{5}) \frac{1}{(1-2z)(5z-4)} \&= \lim_{z \to \frac{4}{5}} \frac{(z - \frac{4}{5})}{(1-2z)(5z-4)} \&= \lim_{z \to \frac{4}{5}} \frac{1}{1-2z} \cdot \lim_{z \to \frac{4}{5}} \frac{z-\frac{4}{5}}{5z-4} \&= \frac{1}{1-2(\frac{4}{5})} \cdot \lim_{z \to \frac{4}{5}} \frac{1}{5} \&= \frac{1}{1-\frac{8}{5}} \cdot \frac{1}{5} \&= -\frac{5}{3}.\end{align*}\]

Key concepts

Complex Analysis

Complex analysis is a branch of mathematics focusing on functions of a complex variable. A complex number has both a real and an imaginary part, written as \(z = x + iy\). Analysts in this field study various properties and applications of such functions. This area of mathematics is essential for solving problems that involve complex functions, like the one in our exercise. Some key concepts include:

• Analytic functions: Functions that are differentiable at every point in their domain.
• Contour integration: Integrating functions along a certain path in the complex plane.
• Holomorphic functions: Another term for analytic functions.

Understanding these concepts provides a solid foundation for dealing with more complex problems, such as finding residues at singular points.

Singularities

Singularities are points where a function is not defined or not analytic. In complex analysis, identifying singularities is crucial because they help determine the behavior of functions. There are different types of singularities:

• Removable singularities: Points where a function appears to be undefined but can be made analytic by an appropriate assignment.
• Poles: Points where a function goes to infinity. The order of the pole refers to how fast the function grows as it approaches the singularity.
• Essential singularities: Points where the function has more chaotic behavior near the singularity.

In our original problem, the function has singularities at \(z = \frac{1}{2}\) and \(z = \frac{4}{5}\). These are identified by solving the denominator of the given function, setting \(1-2z = 0\) and \(5z-4 = 0\).

Residue Theorem

The residue theorem is a powerful tool in complex analysis. It allows us to evaluate contour integrals by focusing on the singularities inside the contour. The theorem states that for a function \(f(z)\) analytic inside and on some simple closed contour \(C\), except at some isolated singularities, the contour integral of \(f(z)\) around \(C\) is \[ \frac{1}{2\pi i} \oint_{C} f(z) \, dz = \sum \text{Res}(f, a_i) \] where \(\text{Res}(f, a_i)\) are the residues at the singularities within the contour. In practical terms, this allows us to break down complex integrals by focusing on simpler parts: the residues at singular points. For our function, we calculated the residues at \(z = \frac{1}{2}\) and \(z = \frac{4}{5}\) using \[ \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z) \], which simplifies the process significantly.

Poles

Poles are specific types of singularities where the function approaches infinity as it nears the singularity. Understanding poles is essential in residue calculus because residues are linked directly to the nature of the poles. There are different types of poles:

• Simple poles (order 1): Where the function grows like \(\frac{1}{z - a}\).
• Higher-order poles: Where the function grows faster, for example, like \(\frac{1}{(z - a)^n}\).

In our problem, both \(z = \frac{1}{2}\) and \(z = \frac{4}{5}\) are identified as simple poles. This simplifies the residue calculations, using the formula \[ \text{Res}(f, a) = \lim_{z \to a} (z-a) f(z) \]. Recognizing the order of the pole helps choose the right method for calculating residues and solving integrals efficiently.