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If $\omega (\neq 1)$ is a cube root of unity, and $(1+\omega )^{7}=A+B\omega .Then\; (A,B)\; equals$

Option 1)
$(1,0)\;$

Option 2)
$\; (-1,1)\;$

Option 3)
$\; (0,1)\;$

Option 4)
$\; (1,1)$

Answers (2)

best_answer

As we have learned

Cube roots of unity -

$z=\left ( 1 \right )^{\frac{1}{3}}\Rightarrow z=\cos \frac{2k\pi }{3}+i\sin \frac{2k\pi }{3}$

k=0,1,2 so z gives three roots

$\Rightarrow 1,\frac{-1}{2}+i\frac{\sqrt{3}}{2}\left ( \omega \right ),\frac{-1}{2}-i\frac{\sqrt{3}}{2}\left ( \omega^{2} \right )$

- wherein

$\omega=\frac{-1}{2} +\frac{i\sqrt{3}}{2},\omega^{2}=\frac{-1}{2} -\frac{i\sqrt{3}}{2},\omega^{3}=1, 1+\omega+\omega^{2}=0$

$1,\omega,\omega^{2}$ are cube roots of unity.

$(1+\omega )^{7}= (-\omega ^2)^7$

$\left ( \because 1+ \omega +\omega ^2 = 0 \right )$

$=- \omega ^{14}$

$=- \omega ^{2}$

$=+ (1+\omega )$

$\Rightarrow A = 1 , B =1$

Option 1)

$(1,0)\;$

Option 2)

$\; (-1,1)\;$

Option 3)

$\; (0,1)\;$

Option 4)

$\; (1,1)$

Posted by

Himanshu

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