Question

The sum of the roots of equation $z^6+64=0,$ whose real part is positive is

• A. $2\sqrt{3}$
• B. $2$
• C. $4$
• D. $\sqrt{3}$

Answer 1

The correct option is A $2\sqrt{3}$

$z^6=-64=2^6\cdot (-1)$
$z=2.e^{i(2k+1)\frac{\pi }{6}},k=0,1,\cdots ,5$

$\therefore z=2e^{i\frac{\pi }{6}},2e^{i\frac{\pi }{2}},2e^{i\frac{5\pi }{6}},e^{i\frac{7\pi }{6}},2e^{i\frac{3\pi }{2}},2e^{i\frac{11\pi }{6}}$ are the roots of the given equation

$\therefore$ Sum of the roots with $+ve$ real part's is $=2e^{i\frac{\pi }{6}}+2e^{i\frac{11\pi }{6}}$
$\Rightarrow 2\left[\right(\cos \pi /6+i\sin \pi /6)+(\cos \left(11\pi /6\right)+i\sin \left(11\pi /6\right)\left)\right]=$
$\Rightarrow 2\sqrt{3}$