Question

If z² – z + 1 = 0, then the value of ( \left( z^{24} + \frac{1}{z^{24}} \right)^{2} ) is

• A. 24
• B. 48
• C. 0
• D. (24)²

Answer 1

Answer: (B)

48

Answer 2

Sol. z² – z + 1 = 0

z = $\frac{1 \pm i\sqrt{3}}{2}$

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

zⁿ = cos $\frac{n\pi}{3}$ + i sin $\frac{n\pi}{3}$, n = 1, 2, …. 24

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

$\left( z + \frac{1}{z} \right)^{2}$ + $\left( z^{2} + \frac{1}{z^{2}} \right)^{2}$ + $\left( z^{3} + \frac{1}{z^{3}} \right)^{2}$+ …$\left( z^{24} + \frac{1}{z^{24}} \right)^{2}$

= 2² cos² $\frac{\pi}{3}$+ 2² cos²$\frac{2\pi}{3}$+ … 2² cos² $\frac{24\pi}{3}$

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

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

Ž 2 [24 + 0] Ž 48