Explanation

(d)
$\left( \cos{\alpha + i\sin\alpha} \right)^{3/5} = e^{i3/5} = e^{i(2n\pi + 3\alpha)/5}$
$\therefore$ Required product $= e^{\frac{i3\alpha}{5}}.e^{\frac{i(2\pi + 3\alpha)}{5}}.$
$e^{\frac{i(4\pi + 3\alpha)}{5}}.e^{\frac{i(6\pi + 3\alpha)}{5}}.e^{i(8\pi + 3\alpha)/5}$
$= e^{i(4\pi + 3\alpha)}$
$= \cos{(4\pi + 3\alpha) + i\sin{(4\pi + 3\alpha)}}$
$= \cos{3\alpha + i\sin{3\alpha}}$