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

VITEEEVITEEE 2007

Solution:

$\left(\cos\alpha +i \sin\alpha\right)^{\frac{3}{5}} = \left(\cos3\alpha+ i \sin3\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 $