Question

The roots of the equation $z^5+z^4+z^3+z^2+z+1=0$ are given by

• A. $-1$
• B. $-\frac{1}{2}+\frac{i\sqrt{3}}{2}$
• C. $\frac{1}{2}+\frac{i\sqrt{3}}{2}$
• D. $\frac{-1-i\sqrt{3}}{2}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $-1$
B $\frac{1}{2}+\frac{i\sqrt{3}}{2}$
C $-\frac{1}{2}+\frac{i\sqrt{3}}{2}$
D $\frac{-1-i\sqrt{3}}{2}$

$z^5+z^4+z^3+z^2+z+1=0\Rightarrow \frac{z^6-1}{z-1}=0\Rightarrow z\ne 1\&z^6=1=\cos 0+i\sin 0\Rightarrow z={(\cos 0+i\sin 0)}^{\frac{1}{6}}=\cos \frac{2k\pi }{6}+i\sin \frac{2k\pi }{6}\Rightarrow z=\cos \frac{k\pi }{3}+i\sin \frac{k\pi }{3}$
where $k=0,1,2,3,4,5$.
For $k=0$,
$z=1$ but $z\ne 1$
For $k=1$,
$z=\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}=\frac{1+i\sqrt{3}}{2}$
For $k=2$,
$z=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=\frac{-1+i\sqrt{3}}{2}$
For $k=3$,
$z=\cos \frac{3\pi }{3}+i\sin \frac{3\pi }{3}=-1$
For $k=4$,
$z=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=\frac{-1-i\sqrt{3}}{2}$
For $k=4$,
$z=\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}=\frac{1-i\sqrt{3}}{2}$
Hence, all the options A,B,C and D are correct.