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)

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