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
$={}^6{C}_0+{}^6{C}_1(i\sqrt{3})+{}^6{C}_2(i\sqrt{3}{)}^2+{}^6{C}_3(i\sqrt{3}{)}^3$
$+{}^6{C}_4(i\sqrt{3}{)}^4+{}^6{C}_5(i\sqrt{3}{)}^5+{}^6{C}_6(i\sqrt{3}{)}^6$
$=({}^6{C}_0-{}^6{C}_2(3)+{}^6{C}_4(9)-{}^6{C}_6(27))$
$+i\sqrt{3}({}^6{C}_1-3{}^6{C}_3+9{}^6{C}_5)$
$=(1-15\times 3+135-27)+i\sqrt{3}(60-60)$
$=(136-72)+i\sqrt{3}(0)=64$
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) :
$z=1+i\sqrt{3}$
$z=2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$
$\Rightarrow z=2\left(cos\frac{\pi }{3}+isin\frac{\pi }{3}\right)$
$\Rightarrow z=2e^{i\pi /3}$
$\therefore z^6=2^6{e}^{2i\pi }=2^6=64(\because cos2\pi =1)$