# Calculate integral using Cauchy's formula

• skrat
In summary, the integral around a circle with center 2 and radius 2, with the function (z-1)sin(z)/(z^2-2z-3) can be calculated using Cauchy's formula by considering the pole at z=3, resulting in the value of 2sin(3)pi/2. Considering the pole at z=-1 is unnecessary as it is outside the contour.
skrat

## Homework Statement

Let ##K## be a circle ##|z-2|=2##. Using Cauchy's formula calculate ##\int _K\frac{(z-1)\sin(z)}{z^2-2z-3}dz##

## Homework Equations

Cauchy's formula: ##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

## The Attempt at a Solution

It's my first time dealing with this kind of problems so I am not really sure that what I am doing is in fact the right way to do it. I would be really happy if one could check my solution.

##\frac{(z-1)\sin(z)}{z^2-2z-3}=\frac{(z-1)\sin(z)}{(z-3)(z+1)}## therefore ##a_1=3## and ##a_2=-1##.

For ##a_1=3##:

##f(z)=\frac{(z-1)\sin(z)}{z+1}##

##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

##\frac{2\sin(3)}{4}2\pi i=\int _K\frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}dz##

For ##a_2=-1##:

##f(z)=\frac{(z-1)\sin(z)}{z-3}##

##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

##\frac{-2\sin(-1)}{-4}2\pi i=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##

##-\pi i\sin(1)=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##

Since ##K## is a circle with center in 2 and radius 2, point ##a_2=-1## is not inside ##K##. Which means that only integral around ##a_1=3## counts and therefore

##\int _K\frac{(z-1)\sin(z)}{z^2-2z-3}dz=\int _K\frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}dz=\frac{2\sin(3)}{4}2\pi i##.

I HOPE so. Thanks in advance!

You have the basic idea down, but what you wrote below is incorrect.

skrat said:
For ##a_2=-1##:

##f(z)=\frac{(z-1)\sin(z)}{z-3}##
For the given K and the way you chose ##a## and f, you can't say that
$$f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}$$ because function f isn't differentiable everywhere inside the region bounded by K. Therefore, the stuff you wrote below isn't correct:
##\frac{-2\sin(-1)}{-4}2\pi i=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##

##-\pi i\sin(1)=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##
Note also that when you considered the other pole, you said
$$\int _K \frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}\,dz = \frac{2 \sin 3}{4}2\pi i.$$ You also said
$$\int _K \frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}\,dz = -\pi i \sin 1.$$ You have the same integral equal to two different quantities!

When you calculated what the integral equaled by considering the pole at z=3, you were done. The pole at z=-1 is outside the contour, so you don't need to consider it at all.

vela said:
When you calculated what the integral equaled by considering the pole at z=3, you were done. The pole at z=-1 is outside the contour, so you don't need to consider it at all.

I completely agree with that. I only wanted to check if my idea is right, where I completely forgot that ##f## may not be Holomorphic outside ##K##.

So in case I would have two singular points inside given ##K##, than the integral would be a sum of two, or not?

Yes, that's how it will turn out, as you will learn later.

Thank you!

## 1. How do I use Cauchy's formula to calculate an integral?

Cauchy's formula states that for a function f(z) that is analytic on a simple closed curve C, the integral of f(z) along C is equal to 2πi times the sum of the residues of f(z) at all singular points inside C. To use this formula, first identify the singular points within the curve C and calculate their residues. Then, plug these values into the formula and solve for the integral.

## 2. Can Cauchy's formula be used to calculate any type of integral?

No, Cauchy's formula can only be used to calculate integrals of analytic functions on simple closed curves. It cannot be used for integrals that involve other types of functions, such as trigonometric or logarithmic functions.

## 3. What is a residue in the context of Cauchy's formula?

A residue is the value of a function at a singular point, multiplied by a specific factor. It is used in Cauchy's formula to calculate the integral of a function along a closed curve. The residue at a singular point is typically found by taking the limit of the function as z approaches the singular point.

## 4. Can Cauchy's formula be used for integrals in the complex plane?

Yes, Cauchy's formula can be used to calculate integrals in the complex plane. It is specifically designed for integrals along simple closed curves in the complex plane, and can be a useful tool in complex analysis.

## 5. How do I know if a function is analytic on a given curve?

A function is considered analytic on a curve if it is continuous and differentiable at all points within the curve. This means that the function must have a well-defined derivative at every point within the curve. Functions that are not analytic, such as those with discontinuities or non-differentiable points, cannot be evaluated using Cauchy's formula.

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