If 1- - and -2 are the cube roots of unity- prove that a-b- -c-2c-a- -b-2 - -2

Given, a+bω +cω^2c+aω +bω^2 =ω^2(a+bω +cω^2)ω^2(c+aω +bω^2) =ω^2(a+bω +cω^2)cω^2+aω^3+bω^4 =ω^2(a+bω +cω^2)(cω^2+a+bω) #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Cube Root of a Complex Number

If 1, ω and...

Question

If $1, ω$ and $ω^{2}$ are the cube roots of unity, prove that $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}} = ω^{2}$

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Solution

Given, $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}$

$= \frac{ω^{2} (a + b ω + c ω^{2})}{ω^{2} (c + a ω + b ω^{2})}$

$= \frac{ω^{2} (a + b ω + c ω^{2})}{c ω^{2} + a ω^{3} + b ω^{4}}$

$= \frac{ω^{2} (a + b ω + c ω^{2})}{(c ω^{2} + a + b ω)}$ $(∵ ω^{3} = 1)$

$= ω^{2}$

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Similar questions

Q. If cube root of unity is

1, ω, ω^{2}

then find the value of:

\frac{(a + b ω + c ω^{2})}{(c + a ω + b ω^{2})} + \frac{(a + b ω + c ω^{2})}{(b + c ω + a ω^{2})}

Q. If

1, ω, ω^{2}

are cube roots of unity and

x = a + b

y = a ω^{2} + b ω, z = a ω + b ω^{2}

, then

x^{2} + y^{2} + z^{2} =

\frac{(a + b ω + c ω^{2})}{c + a ω + b ω^{2}} = ω^{2}

If $ω$ is a complex cube root of unity, then the value of $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}$ + $\frac{a + b ω + c ω^{2}}{b + c ω + b ω^{2}}$ is :

P r o v e t h a t (\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}) = ω^{2}, w h e r e ω i s t h e c u b e r o o t o f u n i t y

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If 1,ω and ω2 are the cube roots of unity, prove that a+bω+cω2c+aω+bω2=ω2

Given, a+bω+cω2c+aω+bω2=ω2(a+bω+cω2)ω2(c+aω+bω2)=ω2(a+bω+cω2)cω2+aω3+bω4=ω2(a+bω+cω2)(cω2+a+bω) (∵ω3=1)=ω2

If $1, ω$ and $ω^{2}$ are the cube roots of unity, prove that $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}} = ω^{2}$

Given, $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}$

$= \frac{ω^{2} (a + b ω + c ω^{2})}{ω^{2} (c + a ω + b ω^{2})}$

$= \frac{ω^{2} (a + b ω + c ω^{2})}{c ω^{2} + a ω^{3} + b ω^{4}}$

$= \frac{ω^{2} (a + b ω + c ω^{2})}{(c ω^{2} + a + b ω)}$ $(∵ ω^{3} = 1)$

$= ω^{2}$