Question

If $z+\frac{1}{z}=\sqrt{3}$ then ${\sum }_{r=1}^5{(z^r+\frac{1}{z^r})}^2$ =

• A. 8
• B. 10
• C. 12
• D. 15

Answer 1

The correct option is A

8

Let z = $rcos\theta$

$z+\frac{1}{z}=\sqrt{3}$

$\Rightarrow (r+\frac{1}{r})cos\theta =\sqrt{3}$

$(r-\frac{1}{r})sin\theta =0\Rightarrow \theta \ne 0$ and r = 1

$\Rightarrow \theta =\frac{\pi }{6}$

$\Rightarrow {(z^n+\frac{1}{z^n})}^2=4co{s}^2\frac{\pi r}{6}$

$\sum {(z^2+\frac{1}{z^n})}^2=4(co{s}^2\frac{\pi }{6}+co{s}^2\frac{\pi }{3}+co{s}^2\frac{\pi }{2}+co{s}^2\frac{2\pi }{3}+co{s}^2\frac{5\pi }{6})=8$