# Contour Integrals: Working Check

• WWCY
In summary, the student is trying to solve two homework problems, but is having trouble with the order of operations and getting confused with the different expressions for odd and even.
WWCY

## Homework Statement

Hi all, could someone help me run through my work for these 2 integrals and see if I'm in the right direction? I'm feeling rather unsure of my work.

1) Evaluate ##\oint _\Gamma Z^*dz## along an anticlockwise circle of radius R centered at z = 0

2) Calculate the contour integral ##\int _C z^n dz## where n ∈ N, and C is a semi-circle contour as shown:

## The Attempt at a Solution

1)
The contour is parameterized by ##Z = Re^{i\theta}##, with ##0 <\theta < 2\pi##. Therefore:

##\oint _\Gamma Z^*dz = \int_{\theta _ 1}^{\theta _2} d\theta \frac{dz}{d\theta} f(z(\theta))## and,

##\int_{\theta _ 1}^{\theta _2} d\theta ( Rie^{i\theta}) (Re^{-i\theta})## thus giving

##R^2i\int_{\theta _ 1}^{\theta _2} d\theta = 2\pi R^2 i##

2)

I define another integral ##I' = \oint_\Gamma Z^n dz## where ##\Gamma## forms a closed loop of the semi-circle with original contour C and an additional contour from 0i to -i that I define as C'

I then have,

##I' = \oint_\Gamma Z^n dz = \int_{C'} Z^n dz + \int_{C}Z^n dz ##

According to Cauchy's integral theorem, ##I'## on the whole should give 0 as ##Z^n## is analytic everywhere in ##\Gamma##, leaving me with

##- \int_{C'} Z^n dz = \int_{C}Z^n dz ##

C' is parameterized by ##Z = it## with ##t## ranging from 0 to -1, hence

##- \int_{C'} Z^n dz = \int_{0}^{-1} dt \frac{dz}{dt} f(z(t))##

##- \int_{C'} Z^n dz = \int_{0}^{-1} dt(i)(it^n)##

## \int_{0}^{-1} dt(i)(it^n) = i^{n+1} \int_{0}^{-1} t^n dt = (i^{n+1})[\frac{-1^{n+1}}{n+1}]##

## \int_{C}Z^n dz = - \int_{C'} Z^n dz = (-1) (i^{n+1})[\frac{-1^{n+1}}{n+1}] = (i^{n+1})[\frac{-1^{n+2}}{n+1}]##

this then gives 2 solutions for odd and even ##n##.

## \int_{C}Z^n dz = \frac{-i^{n+1}}{n+1}## for odd ##n##

## \int_{C}Z^n dz = \frac{i^{n+1}}{n+1}## for even ##n##

Help is greatly appreciated!

#### Attachments

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It looks mostly fine, but you really should not write ##(-1)^k## as ##-1^k##. With correct order of operations, exponentiation comes before subtraction and so ##-1^k = -1##.

Also, instead of writing your result as separate expressions for odd and even ##n##, I suggest using that ##-1 = i^{-2}## in order to simplify your expression significantly.

Orodruin said:
Also, instead of writing your result as separate expressions for odd and even ##n##, I suggest using that ##-1 = i^{-2}## in order to simplify your expression significantly.

Hi @Orodruin, thanks for assisting, I will make the necessary adjustments.

With regards to the quote above, I realized that I could re-write the odd n solution as ##\frac{i^{n-1}}{n+1}##, but I can't see how I can combine the answers as per the suggestion. Do you mind elaborating? Thank you!

WWCY said:
##(i^{n+1})[\frac{(-1)^{n+2}}{n+1}]##

and use ##-1 = i^{-2}## instead of trying to start from the separated expressions.

Orodruin said:

and use ##-1 = i^{-2}## instead of trying to start from the separated expressions.

This gives ##\frac{i^{-(n+1)}}{n+1}## for all n, is that right?

You should check your computation by checking if it holds for some particular ##n##. Try it for ##n = 0##.

Whoops, made a ridiculous error. I believe it should've been ##\frac{i^{-(n+3)}}{n+1} ## instead, is that right?

That is correct, although I would multiply it by ##1 = i^4## to get a slightly more aesthetically pleasing form (not having a 3 in there).

Orodruin said:
That is correct, although I would multiply it by ##1 = i^4## to get a slightly more aesthetically pleasing form (not having a 3 in there).

Right, I'll do that as well! Thank you for your assistance and patience.

## 1. What is a contour integral?

A contour integral is a type of line integral that is used to calculate the integral of a function along a continuous curve in the complex plane. It is also known as a path integral, and is denoted by ∫C f(z) dz.

## 2. How do you calculate a contour integral?

To calculate a contour integral, you need to first parameterize the curve C and then substitute the parameterization into the integral. Then, you can use the fundamental theorem of calculus to evaluate the integral.

## 3. What is the purpose of working check in contour integrals?

The purpose of working check in contour integrals is to verify that the integral has been calculated correctly. It involves checking for errors in the parameterization or substitution process, as well as ensuring that the final answer is consistent with the expected result.

## 4. What are some common mistakes when working with contour integrals?

Some common mistakes when working with contour integrals include incorrectly parameterizing the curve, not accounting for singularities or branch cuts, and making errors in the substitution process. It is important to carefully check your work to avoid these mistakes.

## 5. Are contour integrals only used in complex analysis?

No, contour integrals can also be used in other areas of mathematics such as vector calculus and differential equations. They are also used in physics and engineering to solve problems involving electric fields, fluid flow, and other physical systems.

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