Question

If $z=e^{2\pi i/3}$, then $1+z+3z^2+2z^3+2z^4+3z^5$ is equal to

• A. $-3e^{\pi i/3}$
• B. $3e^{\pi i/3}$
• C. $3e^{2\pi i/3}$
• D. $-3e^{2\pi i/3}$
• E. $0$

Answer 1

The correct option is A $-3e^{\pi i/3}$

$z=e^{2\pi i/3}=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}=\omega$

Therefore, $1+z+3z^2+2z^3+2z^4+3z^5$ $=1+\omega +3{\omega }^2+2{\omega }^3+2{\omega }^4+3{\omega }^5$

$=1+\omega +3{\omega }^2+2+2\omega +3{\omega }^2=3(1+\omega +{\omega }^2)+3{\omega }^2=3{\omega }^2[\because 1+\omega +{\omega }^2=0]=3(-\frac{1}{2}-\frac{\sqrt{3}}{2})=-3(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})=-3e^{\pi i/3}$