'The Road to Reality' Internet Forum - View topic

Page 1 of 1

[ 7 posts ]

Exercise [07.01]
Author	Message
Sameed Zahoor Joined: 12 Mar 2008, 10:57 Posts: 69 Location: India	Exercise [07.01] Let $I=\oint{z^n}dz$ where n is an integer. If n is not equal to -1 then, $I=\frac{z^{(n+1)}(r,\theta+2\pi)}{n+1}-\frac{z^{(n+1)}(r,\theta)}{n+1}$ $=\frac{1}{n+1}(r^{(n+1)}e^{(i\theta+i2\pi)(n+1)}}-r^{(n+1)}e^{(i\theta)(n+1)})$ $=\frac{r^{(n+1)}e^{(i\theta)(n+1)}}{n+1}(e^{i2\pi(n+1)}-1)=0$ If n=-1, $I=\oint{\frac{1}{z}}dz=2\pi\iota$
13 May 2008, 11:11

dacrespi Joined: 28 Mar 2008, 21:15 Posts: 2	Re: Exercise [7.1] I m new in complex analysis. Thank you very much!! Diego
20 May 2008, 22:23

jamestzara Joined: 09 Jun 2008, 09:50 Posts: 2	Re: Exercise [07.01] Great! I wonder, though, seeing as this exercise is marked as "very straight forward" and therefore easily obtainable to the layman, whether an alternative solution would be to note that since there is no ambiguity in raising complex numbers to the power of an integer, the only available contour is one which traverses positively from a to b and then back again, cancelling itself out and yielding zero? The exercise seems to require something that "explains" the result rather than just showing it.
09 Jun 2008, 10:37

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	Re: Exercise [07.01] Yes I agree and here is a simpler way to see it. Remember that is a single-valued function of , unlike $\log z$ , which is multi-valued. $\int_a^bz^ndz=\frac{z^{n+1}}{n+1}\ \|_a^b$ where and are any two points in the complex plane and $n\ne-1$ $=\frac{b^{n+1}}{n+1}-\frac{a^{n+1}}{n+1}$ For a closed loop and because is single-valued the integral vanishes and can be written as $\oint z^ndz=0\ n\ne-1$
09 Jul 2008, 15:41

Sameed Zahoor Joined: 12 Mar 2008, 10:57 Posts: 69 Location: India	Re: Exercise [07.01] vasco wrote: Yes I agree and here is a simpler way to see it. Remember that is a single-valued function of , unlike $\log z$ , which is multi-valued. $\int_a^bz^ndz=\frac{z^{n+1}}{n+1}\ \|_a^b$ where and are any two points in the complex plane and $n\ne-1$ $=\frac{b^{n+1}}{n+1}-\frac{a^{n+1}}{n+1}$ For a closed loop and because is single-valued the integral vanishes and can be written as $\oint z^ndz=0\ n\ne-1$ You are correct,Vasco. However,the reason I wrote the solution in polar-form was to extend the domain of its applicability.The problem specifically mentions 'n+1' to be an integer.But by using polar forms one can also regard 'n+1' as rational or a surd. In such a case, $e^{2\pi\iota\\ (n+1)}-1$ is not zero.This treatment will help to understand more about Riemann Surfaces which have been analysed in the next chapter.For example,if n+1=p/q then the integral vanishes only when 'p' is an integral multiple of q.So the corresponding Riemann Surface is a spiral ramp in which the spiral sheets rejoin after 'q' turns.
10 Jul 2008, 07:13

tensor Joined: 30 May 2009, 19:48 Posts: 2	Re: Exercise [07.01] I think your solution Sameed is the most straightforward, but I'm uncomfortable with you converting into polar form after you have integrated. What's your justification for not substituting $\theta$ with the line integral's limits? Why are you including it?
16 Jul 2009, 23:36

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	Re: Exercise [07.01] I think the point is that in the contour integral you start at a point $z(r,\theta)$ and then go round once back to the point , but $\theta$ has now increased by $2\pi$ so the limits of the integral are $(r,\theta)$ and ( $r,\theta+2\pi)$ .
28 Jul 2009, 18:46

Display posts from previous: Sort by

Page 1 of 1

[ 7 posts ]

Exercise [07.01]

Who is online