Question

If $z^2-z+1=0$, then the value of ${(z+\frac{1}{z})}^2+{(z^2+\frac{1}{z^2})}^2+{(z^3+\frac{1}{z^3})}^2+...+{(z^{24}+\frac{1}{z^{24}})}^2$ is equal to

• A. $44$
• B. $38$
• C. $48$
• D. None of these

Answer 1

The correct option is C $48$

We have, $z^2-z+1=0\Rightarrow z=\frac{1\pm i\sqrt{3}}{2}$

Taking $z=\frac{1+i\sqrt{3}}{2}=\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$

$\Rightarrow z^n=\cos \frac{n\pi }{2}+i\sin \frac{n\pi }{3},n=1,2,...,24$

$\therefore z^n+\frac{1}{z^n}=2\cos \frac{n\pi }{3}$

$\therefore {(z+\frac{1}{z})}^2+{(z^2+\frac{1}{z^2})}^2+{(z^3+\frac{1}{z^3})}^2+...+{(z^{24}+\frac{1}{z^{24}})}^2$

$=2^2{\cos }^2\frac{\pi }{3}+2^2{\cos }^2\frac{2\pi }{3}+2^2{\cos }^2\frac{3\pi }{3}+...+2^2{\cos }^2\frac{24\pi }{3}$

$=2[(1+\cos \frac{2\pi }{3})+(1+\cos \frac{4\pi }{3})+(1+\cos \frac{6\pi }{3})+...+(1+\cos \frac{48\pi }{3})]$

$=2\left[\frac{24+\cos (\frac{2\pi }{3}+\frac{23\pi }{3})\sin \frac{24\pi }{3}}{\sin \frac{\pi }{3}}\right]=48$