Answer to Question #116902 in Algebra for kay

Question

Given that w denotes either one of the non-real roots of the equation z^3 = 1, show
that
(a) 1 + w + w^2 = 0, and
(b) the other non-real root is w^2
. Show that the non-real roots of the equation

((1 − u)÷u)^3
can be expressed in the form Aw and Bw^2
, where A and B are real
numbers, find A and B.

Answer 1

$z^3=1\\ z=\sqrt[3]{1}\\ 1=\cos0+i\sin0\\ \sqrt[3]{1}=\cos\frac{0+2\pi k}{3}+i\sin\frac{0+2\pi k}{3}, k=0,1,2\\ w_0=\cos0+i\sin0=1\\ w_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\frac{1}{2}+i\frac{\sqrt3}{2}\\ w_2=\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}=-\frac{1}{2}-i\frac{\sqrt3}{2}\\$

a)

$1+w_1+w^2_1=1-\frac{1}{2}+i\frac{\sqrt3}{2}+\\ +(-\frac{1}{2}+i\frac{\sqrt3}{2})^2=\frac{1}{2}+i\frac{\sqrt3}{2}+\\ +\frac{1}{4}-i\frac{\sqrt3}{2}-\frac{3}{4}=0\\ 1+w_2+w^2_2=1-\frac{1}{2}-i\frac{\sqrt3}{2}+\\ +(-\frac{1}{2}-i\frac{\sqrt3}{2})^2=\frac{1}{2}-i\frac{\sqrt3}{2}+\\ +\frac{1}{4}-i\frac{\sqrt3}{2}-\frac{3}{4}=0$

b)

$w_1^2=(-\frac{1}{2}+i\frac{\sqrt3}{2})^2= \frac{1}{4}-i\frac{\sqrt3}{2}-\frac{3}{4}=\\ =-\frac{1}{2}-i\frac{\sqrt3}{2}=w_2\\ w_2^2=(-\frac{1}{2}-i\frac{\sqrt3}{2})^2= \frac{1}{4}+i\frac{\sqrt3}{2}-\frac{3}{4}=\\ =-\frac{1}{2}+i\frac{\sqrt3}{2}=w_1$

$(\frac{1-u}{u})^3=1\\ u=Aw_1=A(-\frac{1}{2}+i\frac{\sqrt3}{2})\\ \frac{1-u}{u}=\sqrt[3]{1}\\ \frac{1-u}{u}=w_1\\ \frac{1-Aw_1}{Aw_1}=w_1\\ 1-Aw_1=Aw_1^2\\ A(w_1+w_1^2)=1\\ w_1^2=(-\frac{1}{2}+i\frac{\sqrt3}{2})^2=\\ =\frac{1}{4}-i\frac{\sqrt3}{2}-\frac{3}{4}=\\ =-\frac{1}{2}+i\frac{\sqrt3}{2}=w_2\\ A(w_1+w_2)=1\\ A(-1)=1\\ A=-1$

$u=Bw_2=B(-\frac{1}{2}-i\frac{\sqrt3}{2})\\ 1-Bw_2=Bw_2^2\\ B(w_2+w_1)=1\\ B=-1$