Question

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

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

Answer 1

Answer: (D)

$e^{2\pi i/3}=cos\frac{2\pi }{3}+\frac{isin2\pi }{3}=\omega$
Now, $1+z+3z^2+2z^3+2z^4+3z^5$
$=1+\omega +3{\omega }^2+2{\omega }^3+2{\omega }^4+3{\omega }^5$
$=1+\omega +3{\omega }^2+2+2\omega +3{\omega }^2$
$=3+3\omega +6{\omega }^2=3\left(1+\omega +{\omega }^2\right)+3{\omega }^2$
$=3{\omega }^2=3e^{4\pi /3}$
$=3\left(cos\frac{4\pi }{3}+isin\frac{4\pi }{3}\right)$
$=3\left[cos\left(\pi +\frac{\pi }{3}\right)+isin\left(\pi +\frac{\pi }{3}\right)\right]$
$=3\left[-cos\frac{\pi }{3}isin\frac{\pi }{3}\right]=-3e^{i\pi /3}$