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
w=reiq
where r and qare real (and r is non-negative). Now
w* = r.e-iq = r2 .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 (wn)=1. Therefore w=e2p/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.
w=e2k pi /n
for integer k, 0 £ 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