Question

If $z = 1 + i\sqrt{3}$, then $z^6$ equals

• A. $32$
• B. $- 32$
• C. $64$
• D. None of these

Answer 1

Answer: (C)

$z = 1 + i\sqrt{3}$
$\therefore z^6 = (1 + i \sqrt{3})^6 ...(i)$
Using binomial theorem, (i) becomes
$= \,{}^6C_0 + \,{}^6C_1(i\sqrt{3}) + \,{}^6C_2( i\sqrt{3})^2 + \,{}^6C_3(i\sqrt{3})^3$
$+ \,{}^6C_4(i\sqrt{3})^4 + \,{}^6C_5(i\sqrt{3})^5 + \,{}^6C_6(i\sqrt{3})^6$
$= (\,{}^6C_0 - \,{}^6C_2(3) + \,{}^6C_4(9) - \,{}^6C_6(27))$
$+ i\sqrt{3} (\,{}^6C_1 - 3\,{}^6C_3 + 9\,{}^6C_5)$
$= ( 1 - 1 5 \times 3 + 135 - 27) + i\sqrt{3} (60 - 60)$
$= (136 - 72) + i\sqrt{3} (0) = 64$
Short Cut Method (i) :
$z = 1 + i\sqrt{3} = - \left(\frac{-1 - i\sqrt{3}}{2}\right) \times 2 = - 2 \omega^2$
$ \therefore z^6 = (-2)^6(\omega^2)^6 = 2^6 \omega^{12}$
$2^6 = 2^6 = 64$
Short Cut Method (ii) :
$z = 1 + i\sqrt{3}$
$ z = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) $
$\Rightarrow z = 2\left(cos \frac{\pi}{3} + i\,sin \frac{\pi}{3}\right)$
$\Rightarrow z = 2e^{i\pi/3}$
$\therefore z^6 = 2^6 \,e^{2i\pi} = 2^6 = 64 \,\,(\because cos\,2\pi = 1)$