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

Short answer

The residue of $\frac{z}{(z^2+1{)}^2}$ at $z=i$ is $\frac{1}{4i}$.

Step-by-step solution

Identify the singularity

The given function is $\frac{z}{(z^2+1{)}^2}$. First, identify the singularity at the indicated point $z=i$. Notice that $z^2+1=0$ at $z=i$, so there is a singularity at $z=i$.

Determine the order of the pole

To determine the order of the pole at $z=i$, consider $(z-i{)}^k$. The term $(z^2+1)$ is zero at $z=i$. Because $(z^2+1{)}^2=0$ at $z=i$, it is a pole of order 2.

Calculate the residue using the formula for a pole of order 2

For a pole of order 2, the residue is given by the formula $\text{Res}(f,i)=\frac{1}{1!}\frac{d}{dz}[(z-i{)}^{2}f(z)]{|}_{z=i}$. Substitute $\frac{z}{(z^2+1{)}^2}$ into the formula: $f(z)=\frac{z}{(z-i{)}^2(z+i{)}^2}$. Simplify the function by expanding it around the point $z=i$.

Differentiate and evaluate

Differentiate the function with respect to $z$: $\frac{d}{dz}[(z-i{)}^2\frac{z}{(z^2+1{)}^2}]{|}_{z=i}$. This simplifies to $\frac{d}{dz}[\frac{z}{(z+i{)}^2}]{|}_{z=i}$. Differentiate: $\text{Res}(f,i)=\frac{d}{dz}[\frac{z}{(z+i{)}^2}]{|}_{z=i}=-\frac{1}{(2i)(2i)}{|}_{z=i}=-\frac{1}{-4i}=\frac{1}{4i}$

Key concepts

complex analysis

Complex analysis is the study of functions that operate on complex numbers. It's a fascinating branch of mathematics with roots in solving complex equations and understanding complex functions. One key idea in complex analysis is the function of a complex variable, denoted generally as f(z), where z is a complex number.
Complex analysis is distinct from real analysis due to the presence of the imaginary unit i, which satisfies the equation i² = -1. This allows for rich properties and behaviors not seen in the real numbers.
In the given exercise, we deal with the function $\frac{z}{(z^2+1{)}^2}$ and focus on its behavior near the complex number z=i. This brings us to dealing with 'singularities', a core topic within complex analysis.

singularities

Singularities are points where a function does not behave normally, such as where it becomes infinite or undefined. In complex analysis, singularities are very important because they tell us a lot about the function's behavior.
There are different types of singularities:

• Removable singularities
• Poles
• Essential singularities

In our exercise, we are focusing on a pole. Specifically, at the point z=i for the function $\frac{z}{(z^2+1{)}^2}$. To identify a singularity, we check where the denominator becomes zero because this is where the function could potentially go to infinity or become undefined. For our function, setting the denominator (z^2 + 1)^2 to zero shows us that z = i and z = -i are singularities.
Understanding singularities helps us in calculating important properties like residues, which are used in evaluating complex integrals.

poles

Poles are a special type of singularity where a function goes to infinity in a specific manner. Poles can be of different 'orders', determining how fast the function goes to infinity as we approach the singularity.
In our problem $\frac{z}{(z^2+1{)}^2}$ at z=i, we identified that there is a pole of order 2. This means as z approaches i, the function tends towards infinity at a rate proportional to $1/(z-i{)}^2$.
To find the residue at a pole of order 2, we use a specific formula:
\text{\text{{Res}}(f, i) = \frac{1}{1!} \frac{d}{dz} \big[(z - i)^2 f(z)\big] \big|_{z = i}\}.
Applying this to our function, substituting it into the formula, and differentiating gives us the residue, which is $1/4i$. Calculating residues is crucial in many areas of complex analysis, particularly in evaluating contour integrals using the residue theorem.