Question

The product of all values of $(\cos \alpha +i\sin \alpha {)}^{3/5}$ is equal to

• A. 1
• B. $\cos \alpha +i\sin \alpha$
• C. $\cos 3\alpha +i\sin 3\alpha$
• D. $\cos 5\alpha +i\sin 5\alpha$

Answer 1

Answer: (C)

${\left(\cos \alpha +i\sin \alpha \right)}^{\frac{3}{5}}={\left(\cos 3\alpha +i\sin 3\alpha \right)}^{\frac{1}{5}}$
$={\left[\cos \left(2k\pi +3\alpha \right)+i\sin \left(2k\pi +3\alpha \right)\right]}^{\frac{1}{5}}$
$=\left[\cos \left(\frac{2k\pi +3\alpha }{5}\right)+i\sin \left(\frac{2k\pi +3\alpha }{5}\right)\right],$
where k = 0, 1,2,3,4
Product of all values.
$=\left(\cos \frac{3\alpha }{5}+i\sin \frac{3\alpha }{5}\right).\left(\cos \left(\frac{2\pi +3\alpha }{5}\right)+i\sin \left(\frac{2\pi +3\alpha }{5}\right)\right).\left(\cos \frac{4\pi +3\alpha }{5}+i\sin \frac{4\pi +3\alpha }{5}\right).\left(\cos \frac{6\pi +3\alpha }{5}+i\sin \frac{6\pi +3\alpha }{5}\right).\left(\cos \left(\frac{8\pi +3\alpha }{5}\right)+i\sin \left(\frac{8\pi +3\alpha }{5}\right)\right)$
$=\cos \left[\frac{3\alpha }{5}+\frac{2\pi +3\alpha }{5}+\frac{4\pi +3\alpha }{5}+\frac{6\pi +3\alpha }{5}+\frac{8\pi +3\alpha }{5}\right]+i\sin \left[\frac{3\alpha }{5}+\frac{2\pi +3\alpha }{5}+\frac{4\pi +3\alpha }{5}+\frac{6\pi +3\alpha }{5}+\frac{8\pi +3\alpha }{5}\right]$
$=\cos \left[\frac{5}{2}\left(2.\frac{3\alpha }{5}+\left(5-1\right).\left(\frac{2\pi }{5}\right)\right)\right]+i\sin \left[\frac{5}{2}\left(2.\frac{3\alpha }{5}+\left(5-1\right)..\left(\frac{2\pi }{5}\right)\right)\right]$
$=\cos \left[\frac{5}{2}.\left(\frac{6\alpha }{5}+\left(\frac{8\pi }{5}\right)\right)\right]+i\sin \left[\frac{5}{2}\left(\frac{6\alpha }{5}+\frac{8\pi }{5}\right)\right]$
$=\cos (3\alpha +4\pi )+i\sin (3\alpha +4\pi )$
$=\cos (4\pi +3\alpha )+i\sin (4\pi +3\alpha )$
$=\cos 3\alpha +i\sin 3\alpha$