complex number?

Question

If 1, w and v are the cube roots of unity prove that:
(a) w = conjugate v = v^2
(b) w + v = -1
(c) (w)(v) = 1

thanks

Answer 1

All roots are equiangular from the next root in a polar representation of complex numbers. In the case of three roots, we have a 120° separation. If one root is at 1, the other two roots are a 120° in each direction. That places them at -0.5 ± i½√3. These two polar, complex solutions are conjugates of one another.

If w and v are conjugates then w = a+bi and v = a - bi

w + v = 2a.... not -1.
But, in this particular example, a equals -0.5. So, w + v would in fact be -1

w⋅v = (a + bi)⋅(a - bi) = a² + b²..... not 1.
In this example, however, a = -0.5 and b = ½√3.
a² + b² = (-0.5)² + (½√3)² = 0.25 + 0.75 = 1

w = conj v... this is true... but does w = v² ?
v² = (a - bi)² = (a - bi)⋅(a - bi) = a² - b² - 2abi =
(-0.5)² - (½√3)² - 2(-0.5)(½√3)i = 0.25 - 0.75 + ½i√3 =
-0.5 + ½i√3. This works too.

A good rule to remember, if you dont already know it:
|z|² = z ⋅ ~z

Answer 2

Numbers are 1 , w = 1 /_120° , v = 1 /_240°
Part a)
conj v = 1 /_ (- 240)° = 1 /_120°
w = 1 /_120°
v ² = 1 /_480° = 1 /_120°
as required

Part b)
1 / _120° + 1 /_240°

cos 120° + i sin120° + cos 240° + i sin 240°
- 1/2 - 1/2 + i sin 60° - i sin60°
- 1

Part c)
w v = 1 /_120° 1 /_240°
w v = 1 /_360° = 1 (cos 360° + i sin 360°)
w v = 1

Answer 3

they follow easily from Viete's formulas because they are roots of:

x^3-1=0

Answer 4

If you change the numbers to polar form, you can prove it easily using de Moivre's formula.