Question

The root of the equation $z^5+z^4+z^3+z^2+z+1=0$ having the least positive argument is

• A. $\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$
• B. $\cos \frac{\pi }{5}+i\sin \frac{\pi }{5}$
• C. $\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}$
• D. $\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$

Answer 1

Answer: (A)

$\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$

$z^5+z^4+z^3+z^2+z+1=0$

$z^3(z^2+z+1)+1(z^2+z+1)=0$

$(z^3+1)(z^2+z+1)=0$

$(z+1)(z^2-z+1)(z^2+z+1)=0$

$[z=-1][z=w,w^2]\to$cube unity $cos\frac{\pi }{6}+sin\frac{\pi }{6}$

$z^2-z+1$ never gets a real no solution.