Question

The product of all values of ${\left(\cos \alpha +\iota \sin \alpha \right)}^{\frac{3}{5}}$

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

Answer 1

The correct option is C

$\cos \left(3\alpha \right)+i\sin \left(3\alpha \right)$

Explanation for the correct option:

Given, ${\left(\cos \alpha +\iota \sin \alpha \right)}^{\frac{3}{5}}$
But, ${\left(\cos \alpha +i\sin \alpha \right)}^{\frac{3}{5}}={\left[{\left(\cos \alpha +i\sin \alpha \right)}^3\right]}^{\frac{1}{5}}$

From de Moivre's theorem,
${\left(\cos \alpha +i\sin \alpha \right)}^n=\cos \left(n\alpha \right)+i\sin \left(n\alpha \right)$ where $n$ is an integer

Thus,
${\left[{\left(\cos \alpha +i\sin \alpha \right)}^3\right]}^{\frac{1}{5}}={\left[\cos \left(3\alpha \right)+i\sin \left(3\alpha \right)\right]}^{\frac{1}{5}}$

Also by de Moivre's theorem,
${\left(\cos \alpha +i\sin \alpha \right)}^{\frac{1}{n}}=\cos \left(\frac{2k\mathrm{\pi }+\alpha }{n}\right)+i\sin \left(\frac{2k\mathrm{\pi }+\alpha }{n}\right)$ where $n$ is an integer and $k=0,1,2,\dots ,n-1$

Thus, ${\left[\cos \left(3\alpha \right)+i\sin \left(3\alpha \right)\right]}^{\frac{1}{5}}=\left[\cos \left(\frac{2k\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{2k\mathrm{\pi }+3\alpha }{5}\right)\right],k=0,1,2,3,4$
Thus, the values of ${\left(\cos \alpha +\iota \sin \alpha \right)}^{\frac{3}{5}}$ are $\left[\cos \left(\frac{3\alpha }{5}\right)+i\sin \left(\frac{3\alpha }{5}\right)\right]$, $\left[\cos \left(\frac{2\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{2\mathrm{\pi }+3\alpha }{5}\right)\right]$,$\left[\cos \left(\frac{4\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{4\mathrm{\pi }+3\alpha }{5}\right)\right]$ $\left[\cos \left(\frac{6\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{6\mathrm{\pi }+3\alpha }{5}\right)\right]$ and $\left[\cos \left(\frac{8\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{8\mathrm{\pi }+3\alpha }{5}\right)\right]$
Their product is,

$$\begin{gathered} =\left[\cos \left(\frac{3\alpha }{5}\right)+i\sin \left(\frac{3\alpha }{5}\right)\right]\times \left[\cos \left(\frac{2\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{2\mathrm{\pi }+3\alpha }{5}\right)\right]\times \left[\cos \left(\frac{4\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{4\mathrm{\pi }+3\alpha }{5}\right)\right]\times \left[\cos \left(\frac{6\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{6\mathrm{\pi }+3\alpha }{5}\right)\right]\times \left[\cos \left(\frac{8\mathrm{\pi }+3\alpha }{5}\right)+i\sin \left(\frac{8\mathrm{\pi }+3\alpha }{5}\right)\right] \\ =\left[\cos \left(\frac{3\alpha }{5}+\frac{2\mathrm{\pi }}{5}+\frac{3\alpha }{5}+\frac{4\mathrm{\pi }}{5}+\frac{3\alpha }{5}+\frac{6\mathrm{\pi }}{5}+\frac{3\alpha }{5}+\frac{8\mathrm{\pi }}{5}+\frac{3\alpha }{5}\right)\right]+i\sin \left[\left(\frac{3\alpha }{5}+\frac{2\mathrm{\pi }}{5}+\frac{3\alpha }{5}+\frac{4\mathrm{\pi }}{5}+\frac{3\alpha }{5}+\frac{6\mathrm{\pi }}{5}+\frac{3\alpha }{5}+\frac{8\mathrm{\pi }}{5}+\frac{3\alpha }{5}\right)\right] \\ =\cos \left(\frac{20\mathrm{\pi }}{5}+5\times \frac{3\alpha }{5}\right)+i\sin \left(\frac{20\mathrm{\pi }}{5}+5\times \frac{3\alpha }{5}\right) \\ =\cos (4\mathrm{\pi }+3\mathrm{\alpha })+i\sin (4\mathrm{\pi }+3\mathrm{\alpha }) \\ =\cos \left(3\mathrm{\alpha }\right)+i\sin \left(3\mathrm{\alpha }\right) \end{gathered}$$

Hence option (C) is correct.