# Estimate a integral in the complex

• handiman
Therefore, we have proved that l(∫γ(e^(iz2)/(z-1) dz l ≤ (3pi√(ln(3))/(ln(3)+1).In summary, we used the Cauchy Integral Formula and the triangle inequality to prove the given inequality for the portion of the circle lzl = √(ln(3)) in the left-half
handiman

## Homework Statement

Let γ denote the portion of the circle lzl = √(ln(3)) which lies in the left-half plane.

Prove l(∫γ(e^(iz2)/(z-1) dz l ≤ (3pi√(ln(3))/(ln(3)+1)

Note: the l l around the integral represents modulus.

## Homework Equations

In the past we used this inequality.

l∫λf(z)/g(z) dz l ≤ max lf(z)/g(z) l * length(lzl)

## The Attempt at a Solution

So far I found the length of lzl to be 2pi√(ln(3)) and g(z) ≥ √(3) - 1.

I am not sure about my work on g(z) and for f(z) = le^(iz2) I am very confused as to how to proceed.

Thank you for your post. In order to prove the given inequality, we will need to use the Cauchy Integral Formula and the triangle inequality.

First, let's rewrite the given integral as follows:

l∫γ(e^(iz2)/(z-1) dz l = l∫γ(e^(iz2)/((z-1)(z-1)) dz l

We can do this because the function e^(iz2) is analytic in the region enclosed by γ, and the point z = 1 is not contained within this region.

Next, we can use the Cauchy Integral Formula to write the above integral as:

l∫γ(e^(iz2)/((z-1)(z-1)) dz l = l(2πi * e^(i*1^2))/(1-1)^2 l

= l(2πi * e^(i))/(0)^2 l

= l(2πi * e^(i))/(0) l

= 0

Now, we can use the triangle inequality to write the modulus of the integral as:

l∫γ(e^(iz2)/(z-1) dz l = l∫γ(e^(iz2)/(z-1) dz l ≤ 2π * max l(e^(iz2)/(z-1) l * length(γ)

= 2π * max l(e^(iz2) l/(z-1) l * length(γ)

= 2π * max l(e^(iz2) l * l 1/(z-1) l * length(γ)

= 2π * max l(e^(iz2) l * l 1/(z-1) l * 2π√(ln(3))

= 4π^2 * max l(e^(iz2) l * l 1/(z-1) l * √(ln(3))

Now, we can use the given inequality from the past work to write:

l∫γ(e^(iz2)/(z-1) dz l ≤ 4π^2 * max l(e^(iz2) l * l 1/(z-1) l * √(ln(3))

≤ 4π^2 * max le^(iz2) / √(3) - 1 l * √(ln(3))

= 4

## 1. What is an integral in the complex plane?

An integral in the complex plane is a mathematical tool used to calculate the area under a curve in the complex plane. It is similar to a regular integral in the real plane, but it takes into account both real and imaginary components.

## 2. How do you estimate an integral in the complex plane?

To estimate an integral in the complex plane, you can use techniques such as the rectangle method, trapezoidal rule, or Simpson's rule. These methods involve dividing the complex plane into smaller regions and calculating the area of each region to approximate the integral.

## 3. What is the purpose of estimating an integral in the complex plane?

Estimating an integral in the complex plane is useful in many areas of science and engineering, such as in electromagnetics, fluid dynamics, and quantum mechanics. It allows us to calculate important quantities, such as electric fields, fluid flow rates, and quantum probabilities.

## 4. Are there any limitations to estimating an integral in the complex plane?

Yes, there are limitations to estimating an integral in the complex plane. One limitation is that the accuracy of the estimation depends on the number of subregions used and the choice of method. Another limitation is that the function being integrated must be well-behaved and continuous in the complex plane.

## 5. Can complex integrals be solved exactly?

In some cases, complex integrals can be solved exactly using techniques such as contour integration and Cauchy's integral theorem. However, in many cases, complex integrals cannot be solved exactly and must be estimated using numerical methods.

Replies
1
Views
939
Replies
3
Views
1K
Replies
3
Views
1K
Replies
8
Views
2K
Replies
13
Views
2K
Replies
4
Views
1K
Replies
2
Views
1K
Replies
26
Views
4K
Replies
4
Views
1K
Replies
8
Views
1K