[JEE Main 2006] if z 2 z 1 0 z complex then sum r 1 6 z
If $z^2+z+1=0$ ($z$ complex), then $\sum_{r=1}^{6}\left(z^r+\dfrac1{z^r}\right)^2$ equals
(a) $18$
(b) $54$
(c) $6$
(d) $12$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
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Correct answer: (d) $12$
$z=\omega$, so $z^r+z^{-r}=2\cos\frac{2\pi r}3$. The six squared terms are $1,1,4,1,1,4$, summing to $12$.
JEE Main 2006 · Complex Numbers — verified solution by the Vidaara Team.
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