James Lobo	Posted on Thursday, 08 May, 2003 - 01:00 pm: I was stuck on this question a)Show that if w is an nth root of unity, then the complex conjugate of w is 1/w. b)Deduce that the complex conjugate of ((1-w)^n) is equal to ((w-1)^n). c)Finally, show that ((1-w)^2n) is real. Thanks
Andre Rzym	Posted on Thursday, 08 May, 2003 - 01:12 pm: We can write w as ${\displaystyle w={\mathrm{re}}^{i\theta }}$ where r and $\theta$are real (and r is non-negative). Now ${\displaystyle w*=r.e^{-i\theta }=r^2.1/w}$ so the real issue is to prove that the modulus of w, i.e. r, is equal to 1. By writing down the equation that w satisfies, can you see how to do this? Andre
James Lobo	Posted on Thursday, 08 May, 2003 - 01:54 pm: For a) (to continue what you have said Andre) since w is an nth root of unity, then $(w^n)=1$. Therefore $w=e^{2\pi /n}$. Hence the modulus of w is 1 ie. r=1. Therefore w*=1/w. What would be your solutions to parts b) and c)?
Andre Rzym	Posted on Thursday, 08 May, 2003 - 02:10 pm: We are told w is an nth root of unity. I.e. ${\displaystyle w=e^{2k\pi i/n}}$ for integer k, $0\le k<n$ In other words, you need to acknowledge the multiple values of w. Nevertheless the conclusion is still the same. Somewhat nicer (in my opinion) is to say that wn =1 |wn |=|1| |w|n =1 |w|=1 Andre
Andre Rzym	Posted on Thursday, 08 May, 2003 - 02:29 pm: As for (b), let me do some of it: ((1-w)n )* = ((1-w)* )n ) = (1* -w* )n = (1-1/w)n Can you see how to finish it? Andre
James Lobo	Posted on Thursday, 08 May, 2003 - 03:31 pm: As ((1-1/w)^n)=((w-1)/w)^n. Therefore this is equal to ((w-1)^n)/(w^n) which is equal to ((w-1)^n) as (w^n)=1 What would be your solutions to part c)? Thanks
Andre Rzym	Posted on Thursday, 08 May, 2003 - 04:13 pm: Here's the first bit: (1-w)2n = (1-w)n (1-w)n = (1-w)n (-1*(w-1))n = (-1)n (1-w)n (w-1)n Let me know how you get on ... Andre
James Lobo	Posted on Thursday, 08 May, 2003 - 05:31 pm: Now with ((1-w)^2n)=((-1)^n)((1-w)^n)((w-1)^n) ((1-w)^2n)*=((-1)^n)*((1-w)^n)*((w-1)^n)* =((-1)^n)((w-1)^n)((1-w)^n) This means that the complex conjugate of ((1-w)^2n) is the same result. This means no imaginary terms are present ie. ((1-w)^2n) is real.
Andre Rzym	Posted on Thursday, 08 May, 2003 - 07:57 pm: Looks good to me. Andre