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. $$\frac{e^{2 z}}{4 \cosh z-5} \text { at } z=\ln 2$$

Short answer

The residue at \( z = \ln 2 \) is \( \frac{4}{3} \).

Step-by-step solution

Rewrite the Function in Terms of its Simple Components

Given the function \ \( f(z) = \frac{e^{2z}}{4 \cosh z - 5} \). Rewrite the denominator using the definition of hyperbolic cosine: \ \( \cosh z = \frac{e^z + e^{-z}}{2} \). \ So, \(4 \cosh z = 4 \cdot \frac{e^z + e^{-z}}{2} = 2(e^z + e^{-z}) \).

Substitute and Simplify

Substitute the expression in the denominator: \( f(z) = \frac{e^{2z}}{2(e^z + e^{-z}) - 5} \). \ Simplify this to get: \( f(z) = \frac{e^{2z}}{2e^z + 2e^{-z} - 5} \).

Identify the Singular Points

Identify the points where the denominator is zero: \ \( 2e^z + 2e^{-z} = 5 \). Let \(u = e^z\), we get the equation: \ \( 2u + \frac{2}{u} = 5 \Rightarrow 2u^2 + 2 = 5u \Rightarrow 2u^2 - 5u + 2 = 0 \).

Solve for u and Find z

The quadratic equation \(2u^2 - 5u + 2 = 0\) can be solved using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = -5\), and \(c = 2\): \ \( u = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \). \ So, \(u = 2 \) or \( u = \frac{1}{2} \). Then, \( e^z = 2 \Rightarrow z = \ln 2 \) or \( e^z = \frac{1}{2} \Rightarrow z = -\ln 2 \).

Determine the Residue Using the Formula for Simple Poles

Since we need to find the residue at \( z = \ln 2 \), and we have determined that it is a simple pole: \ The residue of a function \( \frac{g(z)}{h(z)} \) at a simple pole \( z_0 \) is given by: \ \( \text{Residue} = \frac{g(z_0)}{h'(z_0)} \), where \( h(z) = 2e^z + 2e^{-z} - 5 \) and \( g(z) = e^{2z} \). Calculate \(h'(z) \): \ \( h(z) = 2e^z + 2e^{-z} - 5 \Rightarrow h'(z) = 2e^z - 2e^{-z} \). \ At \( z = \ln 2 \), \( e^z = 2 \) and \( e^{-z} = \frac{1}{2} \). Substitute these into \(h'(z)\): \( h'(\ln 2) = 2 \cdot 2 - 2 \cdot \frac{1}{2} = 4 - 1 = 3 \).

Compute the Residue

Finally, calculate the residue: \ \( \text{Residue} = \frac{g(\ln 2)}{h'(z)} = \frac{e^{2 \ln 2}}{3} = \frac{4}{3} \). Thus, the residue of \( \frac{e^{2z}}{4 \cosh z - 5} \) at \( z = \ln 2 \) is \( \frac{4}{3} \).

Key concepts

Complex Analysis

Complex analysis is a branch of mathematics that deals with complex numbers and functions of a complex variable. A complex number has a real part and an imaginary part, usually written as \(z = x + yi\), where \(x\) and \(y\) are real numbers. The power of complex analysis stems from its ability to capture properties of functions and transformations that are not as easily seen in real analysis. Important concepts include analyticity (functions that are complex differentiable), Cauchy-Riemann equations, and contour integration. These are powerful tools used to analyze and solve complex problems that arise in fields such as engineering and physics. Functions in complex analysis can often be represented through power series and Laurent series, which provide invaluable insights into their behavior.

Singularities

In complex analysis, singularities refer to points at which a complex function does not behave nicely in terms of being analytic. There are different types of singularities:

• Removable Singularities: Points where a function is not defined but can be redefined to make it analytic.
• Poles: Points where a function goes to infinity. This is where our exercise focuses on using residues.
• Essential Singularities: Points where the function exhibits very chaotic behavior.

To find a function’s singularity, we can analyze the points where its denominator goes to zero. Understanding these singular points helps in evaluating complex integrals and in other advanced applications like solving differential equations.

Pole Residue Calculation

The residue theorem is a powerful technique in complex analysis used to evaluate contour integrals. The residue at a point is a measure of the contribution of a pole (a type of singularity) to the integral around that point. For a simple pole at \(z_0\), the residue of a function \(f(z) = \frac{g(z)}{h(z)}\) is given by: \[ \text{Residue} = \frac{g(z_0)}{h'(z_0)} \] In our exercise, we saw the function \(f(z) = \frac{e^{2z}}{4 \text{cosh}(z) - 5}\). We identified the poles by setting the denominator equal to zero. Then, we used the formula to calculate the derivative of the denominator function and evaluated it at the pole to find the residue. This approach transforms a potentially complicated integral into a simpler summation of residue contributions, particularly useful in evaluating real integrals and other complex-valued functions.

Quadratic Formula

The quadratic formula is essential in solving second-degree polynomial equations of the form \(ax^2 + bx + c = 0\). It is used in the exercise to find the zeros of the function's denominator, which are crucial for locating singularities. The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In the given exercise, solving \(2u^2 - 5u + 2 = 0\), we identified \(a=2\), \(b=-5\), and \(c=2\). Applying the quadratic formula yielded \(u = 2\) and \(u = \frac{1}{2}\), helping us find the corresponding \(z\) values where the function has poles. This process is not only foundational for solving polynomial equations but also for uncovering critical points in the context of complex functions, enhancing our understanding and ability to analyze such functions effectively.