Question

If w is complex cube root of unity then prove that w2+w3+w4​

Answer 1

Given : w is complex cube root of unity then

To Find  : prove that w2+w3+w4​ = 0

Solution:

w is complex cube root of unity

then 1 . w . w²  = w³ = 1

and  1 + w + w²  = 0

w² + w³ + w⁴

= w²( 1  + w  + w²)

= w² (0)

= 0

Learn More:

a+b)^2+(aw+bw^2)^2+(aw^2+bw)^2=6ab

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Answer 2

$\large\red\longmapsto \bold \red{Given:}$

w is complex cube root of unity then :-

$\large \red \longmapsto \bold \red{To \: \: Find: }$

$\bold \blue{prove \: that \: {w}^{2} + {w}^{3} + {w}^{4} = 0}$

$\red\longmapsto \bold \red{Solution:}$

w is complex cube root of unity

→ Then 1. w. w² = w³ = 1

→ And 1+w+w² = 0

$→\bold{ {w}^{2} + {w}^{3} + {w}^{4} }$

= w²(1+w+w²)

=w² (0)

= 0

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