Author: vasco [ 28 Feb 2009, 09:01 ] Post subject: Exercise [08.04] c Here is a development of the previous proof, which takes into account the special case when the circle in the z-plane passes through the origin and is transformed into a straight line (circle with infinite radius as Penrose calls it).I noticed a small typing error in the attached .pdf file which I have now corrected."Equation (5) is of the same form as equation (1)" should be "Equation (5) is of the same form as equation (2)"Thanks to Azrael's post below I have corrected a typo and added a clarification. See Azrael's post for details. The pdf file below has also been amemded to correct the typo and add the clarification. Edited by Vasco on 10/10/2011.VascoAttachment: RTRex8.04v2.pdf [63.06 KiB] Downloaded 281 times

 Author: cwjian [ 25 Jul 2011, 18:13 ] Post subject: Re: Exercise [08.04] c Vasco, I have a question. Why does shifting the plane by i upwards not work, whereas rotating the plane by pi/2 radians and then subtracting 1 does?

 Author: vasco [ 25 Jul 2011, 21:32 ] Post subject: Re: Exercise [08.04] c HiI am a bit confused! Are you sure you are talking about this exercise? Don't you mean Exercise 8.7?If you do mean this exercise, can you explain in a bit more detail what you mean?ThanksVasco

 Author: cwjian [ 28 Jul 2011, 15:22 ] Post subject: Re: Exercise [08.04] c Sorry Vasco, yes, I posted in the wrong thread...Would you mind if I continued this discussion in the other thread? Or should I continue it here?

 Author: vasco [ 28 Jul 2011, 15:51 ] Post subject: Re: Exercise [08.04] c Hi cwjianI think it would be more useful to everybody if you transferred to the other thread. Presumably the one for exercise 8.7 - Is that right?Also could you explain in a bit more detail what it is that you do not understand?ThanksVasco

 Author: cwjian [ 29 Jul 2011, 18:19 ] Post subject: Re: Exercise [08.04] c Hi Vasco,I've posted in the Problem 8.07 thread.cwjian

 Author: Azrael84 [ 09 Oct 2011, 15:05 ] Post subject: Re: Exercise [08.04] c Nice solution. I believe there are few typos, equation (1) should really look like:z=z_0+re^{it}, but this doesn't affect the proof obviously. Also on the last page I think you mean to take z_0=x_0+iy_0 (missing an i as z_0 is complex constant), again just some typos hopefully make it easier for those going through your proof.

 Author: vasco [ 10 Oct 2011, 07:48 ] Post subject: Re: Exercise [08.04] c Azrael84 wrote:Nice solution. I believe there are few typos, equation (1) should really look like:z=z_0+re^{it}, but this doesn't affect the proof obviously. Also on the last page I think you mean to take z_0=x_0+iy_0 (missing an i as z_0 is complex constant), again just some typos hopefully make it easier for those going through your proof.Thanks for reading my solution. I'm glad you like it!I agree that on the last page it should be z_0=x_0+iy_0, but I don't agree with your first point.r is a complex number here and so z=z_0+r is correct. That's why I called it a radius vector in line 1. I think that I should have added |r|=constant to make it clearer.As z changes so does r, but only the argument of r changes not its modulus.I will amend my first post above and upload a corrected and amended pdf file.

 Author: Azrael84 [ 10 Oct 2011, 19:01 ] Post subject: Re: Exercise [08.04] c vasco wrote:but I don't agree with your first point.r is a complex number here and so z=z_0+r is correct. That's why I called it a radius vector in line 1. I think that I should have added |r|=constant to make it clearer.As z changes so does r, but only the argument of r changes not its modulus.I will amend my first post above and upload a corrected and amended pdf file.Ah, OK. Yes I agree it makes sense if r is complex. I thought you had in mind 'r' as the radial type parameter of polar coords centered about the point z_0 and had accidentally missed off the e^{it} factor...

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