Question

Given that w denotes either one of the non-real roots of the equation z3 = 1, show that
(a) 1 + w + w2 = 0, and (b) the other non-real root is w2. Show that the non-real roots of the equation 
1 − u u
3 numbers, find A and B.
11. Given that 1, w, w2 are the cube roots of unity, find the equation whose roots are 1/3, 1/(2 + w), 1/(2 + w2).
can be expressed in the form Aw and Bw2, where A and B are real

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

$$w_1=1,\ real$$

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

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

b)

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

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

Then

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

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

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

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

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

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

11.

$$w={-1- i\sqrt{3}\over2}=>2+w={3- i\sqrt{3}\over2}=>$$

$$=>{1\over 2+w}={2\over 3-i\sqrt{3}}={2\over 12}(3+i\sqrt{3})={1\over 6}(3+i\sqrt{3})$$

$$w_2={-1+ i\sqrt{3}\over2}=>2+w_2={3+ i\sqrt{3}\over2}=>$$

$$=>{1\over 2+w_2}={2\over 3+i\sqrt{3}}={2\over 12}(3-i\sqrt{3})={1\over 6}(3-i\sqrt{3})$$

$$(z-{1\over 3})\bigg(z-{1\over 6}(3+i\sqrt{3})\bigg)\bigg(z-{1\over 6}(3-i\sqrt{3})\bigg)=0$$

$$(z-{1\over 3})(z^2-z+{1\over 3})=0$$

$$z^3-z^2+{1\over 3}z-{1\over 3}z^2+{1\over 3}z-{1\over 9}=0$$

$$z^3-{4\over 3}z^2+{2\over 3}z-{1\over 9}=0$$

Or

$$9z^3-12z^2+6z-1=0$$