Find Complex NumbersDate: 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 |
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