Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{z}{\left(z^{2}+1\right)^{2}} \text { at } z=i $$

Short answer

The residue of the function \( \frac{z}{(z^{2}+1)^{2}} \) at \( z=i \) is \( \frac{1}{2i} \).

Step-by-step solution

Identify the Singularity

Identify the type of singularity at the given point, which is at \( z=i \). The denominator \( \left(z^{2}+1\right)^{2} \) reveals a pole of order 2 at \( z=i \) because \( z^{2}+1 = (z+i)(z-i) \).

Write the Function in Terms of \( (z-i) \)

Express the function in a form involving \( (z-i) \). Let \( z = i + w \), then \( z^2 + 1 = (i+w)^2 + 1 = -1 + 2iw + w^2 + 1 = w^2 + 2iw \). Thus, \( f(z) = \frac{i+w}{(w^2 + 2iw)^2} \).

Simplify the Function

Simplify \( \frac{i+w}{(w^2 + 2iw)^2} \) as follows: \( f(w) = \frac{i+w}{4iw^2(w/2i + 1)^2} \). The leading term in the expansion of the denominator near \( w = 0 \) will be \( (2iw)^{-2} \).

Expand in Laurent Series

Expand the function as a Laurent series around \( w = 0 \), identifying the residue. The series expansion will highlight terms of type \( \frac{C}{w} \), which allows identification of residue terms involving \( (i + w) \).

Apply the Residue Formula for a Pole of Order 2

For a pole of order 2 at \(z = i\), use the general residue formula for higher-order poles: \[ \text{Res}(f, i) = \frac{1}{1!}\frac{d}{dz}\bigg[ (z - i)^{2} f(z) \bigg] \bigg|_{z = i} \] Calculate \( (z - i)^{2}f(z) = \frac{z}{w^2 4iw} = \frac{i+w}{4iw^3} \), differentiating with respect to \( w \).

Calculate and Simplify

Differentiate the term \( \frac{i+w}{4iw^2} \) to get the coefficient of \( \frac{1}{w} \), thus obtaining the residue. This yields \[ \text{Res}(f, i) = \frac{d}{dw}\bigg[ \frac{i+w}{4iw^2} \bigg] \bigg|_{w = 0} = \frac{1}{2i} \]

Key concepts

poles of complex functions

A pole of a complex function is a type of singularity where the function diverges to infinity. In simpler terms, if we have a function that goes to infinity at a certain point, that point is called a pole. The order of the pole is important as it defines how rapidly the function grows near the singularity.
For instance, in the exercise, the function has a pole at the point \(z=i\). This is because the denominator \((z^2+1)^2\) goes to zero at \(z=i\), causing the function to blow up. The order of this pole is 2 because the denominator squared magnifies the divergence. Understanding the order helps in applying correct techniques, such as Laurent series or residue calculus.

Laurent series expansion

The Laurent series is a representation of a complex function as an infinite series, which includes terms with negative powers. It's used to analyze functions near singularities. Unlike Taylor series, which is valid only for non-singular points, the Laurent series deals with singularities by providing an expansion with terms like \(\frac{1}{(z-z_0)}\).
In the example exercise, we use the Laurent series to expand our function around the point where the pole occurs. This involves expressing the function in simpler terms and finding the coefficients of the terms in the series. Identifying these coefficients helps pin down the residue, a crucial value that simplifies complex integrations around the singularity.

calculus of residues

The calculus of residues is a powerful technique in complex analysis used to evaluate complex integrals. It describes how to handle integrals around points where functions behave badly (singularities). The core idea is that the integral of a function around a closed loop is determined by the sum of residues inside that loop.
The residue at a given point \(z_0\) is essentially the coefficient of the \(\frac{1}{(z-z_0)}\) term in the Laurent series expansion. In the step-by-step solution of our exercise, we use the residue formula for higher-order poles to find the residue at \(i\). Calculating and applying this residue allows for evaluating integrals involving the function.

higher-order poles

Higher-order poles refer to poles where the function diverges more rapidly than a simple pole. For example, a pole of order 2 is one where the function behaves like \(\frac{1}{(z-z_0)^2}\). These require more nuanced techniques to find residues.
The residue formula for higher-order poles, \(\text{Res}(f, z_0) = \frac{1}{(m-1)!} \bigg[ \frac{d^{m-1}}{dz^{m-1}} \bigg( (z-z_0)^m f(z) \bigg) \bigg|_{z=z_0} \bigg]\), gives a precise method for calculating these residues. In the exercise, we use this formula for a second-order pole at \(i\). Differentiating and simplifying the function as specified reveals the residue, showcasing the power of these advanced techniques for higher-order singularities.