# Calc Residue of \frac{z \sin{z}}{\left( z - \pi \right)^3}

• eXorikos
In summary, the residue for the singularity of \frac{z \sin{z}}{\left( z - \pi \right)^3} cannot be calculated using the limit method due to an incorrect answer. However, it can be calculated by integrating around a path, with a correction for the y integrals. It may be easier to calculate the integral by letting x run from 0 to 2 pi to take advantage of the periodicity of sin z.

## Homework Statement

Calculate the residu for the singularity of $$\frac{z \sin{z}}{\left( z - \pi \right)^3}$$

## Homework Equations

$$R \left( a_0 \right) = \frac{1}{2 \pi i} \oint \frac{z \sin{z}}{\left( z - \pi \right)^3} dz$$

## The Attempt at a Solution

$$\pi$$ is an essential singularity so the residue cannot be calculated using the limit.
I have tried calculating the integral around the path from 3-i to 4-i to 4+i to 3+i to 3-i. In principle this should work, but Maple gives me a wrong answer if I try to evaluate.

$$\frac{1}{2 \pi i} \left( \int^4_3 \frac{\left( x - i \right) \sin{x - i}}{\left( x - i - \pi \right)^3} dx + \int^1_{-1} \frac{\left( 4 + iy \right) \sin{4 + iy}}{\left( 4 + iy - \pi \right)^3} dy + \int^3_4 \frac{\left( x + i \right) \sin{x + i}}{\left( x + i - \pi \right)^3} + \int^{-1}_1 \frac{\left( 3 + iy \right) \sin{3 + iy}}{\left( 3 + iy - \pi \right)^3} dy \right)$$

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How did you conclude that $z=\pi$ is an essential singularity?

I was wrong on that. Calculating the residue using the limit works fine, but my problem stands. The integral should give the same result.

You're missing a factor of i on the y integrals since dz = i dy.

It works fine thanks!

Is there a better choice that makes it easier to calculate the integral?

Perhaps let x run from 0 to 2 pi to take advantage of the periodicity of sin z, but I'd try to avoid the contour integrals in the first place.

## 1. What is the definition of a residue in complex analysis?

A residue is the value of the function at a point where it is not defined. In complex analysis, it is used to calculate the integral of a function around a singularity or pole.

## 2. How do you calculate the residue of a function with a simple pole?

To calculate the residue of a function with a simple pole, you need to find the coefficient of the term with the pole in the Laurent series expansion of the function around the pole. This coefficient is the residue. In this case, the function has a simple pole at z = π, so we need to find the coefficient of the \frac{1}{z - \pi} term in the Laurent series.

## 3. What is the formula for calculating the residue of a function with a triple pole?

The formula for calculating the residue of a function with a triple pole is:Res(f, z_0) = \frac{1}{2\pi i}\lim_{z \to z_0} \frac{d^2}{dz^2}(z-z_0)^3f(z)

## 4. How do you calculate the residue of the given function?

To calculate the residue of the given function, we need to first find the Laurent series expansion of the function around the pole at z = π. Then, we can find the coefficient of the \frac{1}{z - \pi} term, which will be the residue.

## 5. What is the significance of calculating the residue of a function?

Calculating the residue of a function is important in complex analysis because it allows us to evaluate complex integrals around singularities or poles. It is also useful in solving differential equations and in finding the values of multiple integrals.