Answer to Question #117023 in Complex Analysis for Alikor hayford

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 Bw2
, where A and B are real
numbers, find A and B

Answer 1

$$z^3=1$$ $$z^3-1=0$$ $$(z-1)(z^2+z+1)=0$$

$z^2+ z+1=0$

$$z={-1\pm\sqrt{1^2-4(1)^2}\over2}={-1\pm i\sqrt{3}\over2}$$

Cube Root of Unity Value

$$w_1=1,\ real$$

$$w_2={-1- i\sqrt{3}\over2}, complex$$

$$w_3={-1+ i\sqrt{3}\over2}, \ complex$$

$$z^3-u^3=0$$

$$z^3-u^3=(z-u)(z^2+u z+u^2)$$

Then

$$(z-u)(z^2+u z+u^2)=0$$

$z^2+u z+u^2=0$

$$z={-u\pm\sqrt{u^2-4u^2}\over2}=u({-1\pm i\sqrt{3}\over2})$$

$$z_1=u w_1=u\cdot1=1\cdot u+i\cdot0, \ real$$

$$z_2=u w_2=u\cdot{-1- i\sqrt{3}\over2}=-{1\over 2}u-i\cdot{u\sqrt{3}\over 2},\ complex$$

$$z_3=u w_3=u\cdot{-1+ i\sqrt{3}\over2}=-{1\over 2}u+i\cdot{u\sqrt{3}\over 2},\ complex$$