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. $$\frac{z^{2}}{z^{4}+16} \text { at } z=\sqrt{2}(1+i)$$

Short answer

The residue of \[ \frac{z^{2}}{z^{4}+16} \] at \ z= \text{√2}(1+i)\ is -Amount due is simplified fraction form

Step-by-step solution

Express the function in a simpler form

The given function is \(\frac{z^{2}}{z^{4}+16}\). Factor the denominator to find the poles. The denominator can be factored as: \[z^4 + 16 = (z^2 + 4i)(z^2 - 4i) = (z^2 + 4i)(z - \root4(-16))(z + \root4(-16))\] where \root4(-16) are the fourth roots of -16.

Identify the poles

The poles are the roots of the denominator which are \(z = e^{i\frac{5\text{π}}{8}}, e^{i\frac{3\text{π}}{8}}, -e^{i\frac{5\text{π}}{8}}, -e^{i\frac{3\text{π}}{8}}\). For \(z = \root4(-16)\), substituting \(z = \text{√2}(1+i)\), we get \[ \text{√2}(1+i) = 2^{1/2} * (1 + i) = \text{root4}(-16) \ Therefore \text{√2}(1+i) is an actual pole.\]

Calculate the residue

To find the residue at \(z = \text{√2}(1+i)\), the formula for the residue at a simple pole is: \(\text{Res}\bigg(f(z), z_0\bigg) = \frac{g(z_0)}{h'(z_0)} \) For the first term after simplified function:Where: \(\frac{z^2}{(z^2-(2e^{-iπ/4}i)} (z^2-(2eiπ/4i))} And at last multiply inside roots, numerator stays z^2\). Simply plug the parts of the function; we get the fraction of parts.

Key concepts

complex analysis

Complex analysis is a fascinating branch of mathematics that studies functions of complex numbers. It involves using complex variables and has many practical applications in physics and engineering. Complex numbers consist of a real part and an imaginary part, which can be written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\).

One fundamental concept in complex analysis is the idea of analytic functions, which are functions that are complex-differentiable in a region. These functions can be represented by power series, like \(f(z) = \text{Σ} c_n (z - z_0)^n\), within their radius of convergence. Analytic functions are very well-behaved and allow for deeper analysis using tools like the Residue Theorem.

The Residue Theorem is a powerful tool in complex analysis. It provides an efficient way to evaluate complex line integrals using residues at poles within a contour. It arises from the Cauchy Integral Theorem and has deep implications for evaluating integrals, especially those that are otherwise challenging to solve.

poles

In complex analysis, a pole of a function is a point where the function takes an infinite value. More formally, a point \(z = z_0\) is a pole of a function \(f(z)\) if \(f(z)\) behaves like \(\frac{1}{(z - z_0)^n}\) near \(z_0\), where \(n\) is a positive integer.

Poles are important in understanding the behavior of functions and are classified based on their order. A pole of order 1 is called a simple pole, while a pole of order 2 is called a double pole, and so on. Identifying poles and their orders is crucial in applying the Residue Theorem.

For instance, consider the function \(\frac{z^2}{z^4+16}\), which has poles where \(z^4 + 16 = 0\). Factoring gives us \(z^4 + 16 = (z^2 + 4i)(z - \sqrt[4]{-16})(z + \sqrt[4]{-16})\). By finding the roots, we identify the poles at \(z = e^{i\frac{5π}{8}}, e^{i\frac{3π}{8}}, -e^{i\frac{5π}{8}}, -e^{i\frac{3π}{8}}\).

residues

The concept of a residue is central to the Residue Theorem. The residue of a function \(f(z)\) at a complex number \(z_0\) is essentially the coefficient of the \((z - z_0)^{-1}\) term in the Laurent series expansion of \(f(z)\) about \(z_0\). This coefficient provides insight into the function's behavior near the pole.

To calculate the residue at a simple pole, you can use the formula: \(\text{Res}(f(z), z_0) = \frac{g(z_0)}{h'(z_0)}\) where \(f(z) = \frac{g(z)}{h(z)}\), and \(h(z)\) has a simple zero at \(z = z_0\). For example, in the function \(\frac{z^2}{z^4 + 16}\), at the pole \(z = \sqrt{2}(1 + i)\), we factor the denominator and find the residue using the given formula.

This calculation helps in solving complex integrals as residues make evaluating integrals along contours significantly more straightforward.

partial fractions

Partial fractions are a method used to decompose complex rational functions into simpler fractions that are easier to integrate or analyze. They are particularly useful in complex analysis when dealing with functions that have multiple poles.

For example, consider the function \(\frac{z^2}{z^4 + 16}\). To find the residues at different poles, we can express the denominator in its factored form and then use partial fraction decomposition. This allows us to isolate the behavior of the function around each pole, simplifying the residue calculation.

By decomposing into partial fractions, each term can be analyzed individually. This step-by-step approach makes it easier to manage even the most complex expressions and is instrumental in applying techniques like the Residue Theorem.