Question

If w is a cube root of unity, then the value of (w+1)/w is *

a) 1

b) -1

c) w

d) -w​

Answer 1

Basic Concept :-

Let we first derive the cube roots of unity.

$\rm :\longmapsto\:Let \: x = {(1)}^{ \frac{1}{3} }$

On cubing both sides, we get

$\rm :\longmapsto\: {x}^{3} = 1$

$\rm :\longmapsto\: {x}^{3} - 1 = 0$

$\rm :\longmapsto\: (x - 1)( {x}^{2} + x + 1) = 0$

$\bf\implies \:x = 1$

Or

$\rm :\longmapsto\: {x}^{2} + x + 1 = 0$

We know,

Quadratic formula,

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

So,

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4(1)(1)} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1} - 4} }{2}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{3} i}{2}$

$\bf\implies \:x = \dfrac{ - 1 + \sqrt{3}i }{2} \: \: or \: \: \dfrac{ - 1 - \sqrt{3} i}{2}$

Hence,

Cube roots of unity are,

$\bf\implies \:1, \dfrac{ - 1 + \sqrt{3}i }{2} \: \: and \: \: \dfrac{ - 1 - \sqrt{3} i}{2}$

in symbolic form represented as

$\rm :\longmapsto\:1, \: \omega, \: { \omega}^{2}$

Having properties,

$\rm :\longmapsto\:1 + \omega + {\omega}^{2} = 0$

$\rm :\longmapsto\:\omega = \dfrac{1}{ {\omega}^{2} }$

$\rm :\longmapsto\: {\omega}^{2} = \dfrac{1}{ {\omega} }$

$\rm :\longmapsto\: {\omega}^{3} = 1$

Let's solve the problem now!!!

$\rm :\longmapsto\:\dfrac{\omega + 1}{\omega}$

$\rm \: = \: \dfrac{\omega}{\omega} + \dfrac{1}{\omega}$

$\rm \: = \: 1 + {\omega}^{2}$

$\rm \: = \: - \omega$

$\bf :\implies\:\red{\boxed{\dfrac{ \bf \: \omega + 1}{\omega} = - \omega}}$

Hence, option (d) is correct.