Find Complex Numbers

Date: 12/16/95 at 22:30:56
From: Anonymous
Subject: Algebra (complex numbers)

   Find all complex numbers such that 
   (conjugate z)(z)^(n-1) = 1.
   z is a complex number. ^ means "to the power of".

Date: 5/30/96 at 14:59:10
From: Doctor Charles
Subject: Re: Algebra (complex numbers)

Polar form helps with this problem. If you write z as r * exp(i*t)
the (conjugate z) = r * exp(-i*t). Then taking the magnitude and
argument of both sides gives two equations, one in r and one in t.
(I assume that n is an integer.)

r^n = 1,        i*(n-2)*t = 2*pi*i*k  where k is any integer.

Notice if n=0 then any r will work, otherwise r=1.
If n=2 then any t works, otherwise
    t=(2*pi*k)/(n-2)

So in summary
n=0   =>   z= x (a real number)
n=2   =>   z= exp (it) for any t

Otherwise the (n-2) possibilities for z are the (n-2)th complex
roots of 1.

-Doctor Charles,  The Math Forum

Date: 5/30/96 at 15:0:12
From: Doctor Anthony
Subject: Re: Algebra (complex numbers)

We could write the equation 
    r.e^(-i.theta).r^(n-1).e^(i(n-1)theta) = 1

This gives r^n.e^{i(n-2)theta} = 1

r=1, theta=0 is one possible solution.  In general we require r=1 and
  e^{i(n-2)theta} = 1  and this could be written
    z^(n-2) = {cos(2k.pi) + i.sin(2k.pi)}  k =0, 1, 2,...(n-3) 

Take the (n-2)th root of both sides and apply DeMoivre's theorem 
to get

      z = cos(2kpi/(n-2)) + i.sin(2kpi/(n-2)}  k = 0, 1, 2 ...(n-3)

-Doctor Anthony,  The Math Forum