# Complex contour integral proof

• GGGGc
GGGGc
Homework Statement
How can I show that from the contour C_N (I’ve attached) that absolute value of cot(pi*z) is less than or equal to 0 everywhere on vertical parts of C_N and less than or equal to a value everywhere on the horizontal parts?
Relevant Equations
|a+b|<=|a|+|b|
|a-b|>=|a|-|b|
I’ve attached my attempt. I’ve tried to use triangle inequality formula to attempt, but it seems I got the value which is larger than 1. Which step am I wrong? Also, it seems I cannot neglect the minus sign in front of e^(N+1/2)*2pi. How can I deal with that?

#### Attachments

• IMG_0099.jpeg
69.2 KB · Views: 45
• IMG_0100.jpeg
61.1 KB · Views: 46
• IMG_0101.jpeg
61.8 KB · Views: 46
You got as far as $$|\cot \pi z | = \left|\frac{e^{2i\pi z} + 1}{e^{2i\pi z} - 1} \right| = \left| \frac{e^{2i\pi x}e^{-2\pi y} + 1}{e^{2i\pi x}e^{-2\pi y} - 1}\right|$$ Why not go further? Multiply numerator and denominator by the complex conjugate of the denominator. Then you can easily write down the exact value of $|\cot \pi z|^2$, and at that point you can start trying to bound it on each side of the contour.

Or write $$|\cot \pi z| = \frac{|e^{2i\pi x} + e^{2\pi y}|}{|e^{2i\pi x} - e^{2\pi y}|}$$ and then you can obtain an upper bound by maximizing the distance between $e^{2i\pi x}$ and $-e^{2\pi y}$ in the numerator and mimizing the distance between $e^{2i\pi x}$ and $e^{2\pi y}$ in the denominator.

Last edited:

## What is a complex contour integral?

A complex contour integral is an integral where the function to be integrated is complex-valued, and the path of integration, known as the contour, is a curve in the complex plane. It generalizes the concept of real integrals to complex functions and is a fundamental tool in complex analysis.

## How do you evaluate a complex contour integral?

To evaluate a complex contour integral, you typically parameterize the contour, substitute this parameterization into the integral, and then integrate with respect to the parameter. Alternatively, if the function is analytic and certain conditions are met, you can use powerful theorems such as Cauchy's Integral Theorem or the Residue Theorem to simplify the evaluation.

## What is Cauchy's Integral Theorem?

Cauchy's Integral Theorem states that if a function is analytic and defined on a simply connected domain, then the integral of the function over any closed contour within that domain is zero. This theorem is a cornerstone of complex analysis and simplifies the evaluation of many complex integrals.

## What is the Residue Theorem?

The Residue Theorem is a powerful result in complex analysis that allows for the evaluation of contour integrals by relating them to the sum of residues of the integrand at its singular points inside the contour. A residue is a complex number that encapsulates the behavior of a function near a singularity.

## Can you provide an example of a complex contour integral proof?

Sure! Consider evaluating the integral of $$f(z) = \frac{1}{z-1}$$ around a closed contour that encircles $$z = 1$$. By Cauchy's Integral Formula, the integral is $$2\pi i$$ times the residue of $$f(z)$$ at $$z = 1$$. The residue here is 1, so the integral evaluates to $$2\pi i \times 1 = 2\pi i$$.

• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
571
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
787
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
32
Views
2K
• Calculus and Beyond Homework Help
Replies
26
Views
4K
• Calculus and Beyond Homework Help
Replies
1
Views
2K