If x-a-y-b- -z-c- 2 where - is a complex cube root of unity- then x a - y b - z c -

The correct option is D 0 xa+yb+zc =aa+bωb+cω^2c =1+ω+ω^2 = 0 #160; #160; ( ∵ 1+ω +ω #160; ^ 2 =0 ) ...

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Standard XII

Mathematics

Fundamental Laws of Logarithms

If x=a,y=bω...

Question

If $x = a, y = b ω, z = c ω^{2}$ where $ω$ is a complex cube root of unity, then $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} =$

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Solution

The correct option is D $0$
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c}$
$= \frac{a}{a} + \frac{b ω}{b} + \frac{c ω^{2}}{c}$
$= 1 + ω + ω^{2}$
$= 0$ $(∵ 1 + ω + ω^{2} = 0)$

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If x=a,y=bω,z=cω2 where ω is a complex cube root of unity, then xa+yb+zc=

The correct option is D 0xa+yb+zc=aa+bωb+cω2c=1+ω+ω2=0 (∵1+ω+ω2=0)

If $x = a, y = b ω, z = c ω^{2}$ where $ω$ is a complex cube root of unity, then $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} =$

The correct option is D $0$
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c}$
$= \frac{a}{a} + \frac{b ω}{b} + \frac{c ω^{2}}{c}$
$= 1 + ω + ω^{2}$
$= 0$ $(∵ 1 + ω + ω^{2} = 0)$