Q44 / 120 · VITEEE 2007 · Mathematics
The product of all values of $(\cos \alpha+\mathrm{i} \sin \alpha)^{3 / 5}$ is equal to
- A1
- B$\cos \alpha+\mathrm{i} \sin \alpha$
- Correct answerC$\cos 3 \alpha+\mathrm{i} \sin 3 \alpha$
- D$\cos 5 \alpha+\mathrm{i} \sin 5 \alpha$
Solution
$\begin{aligned} &(\cos \alpha+i \sin \alpha)^{3 / 5}=(\cos 3 \alpha+i \sin 3 \alpha)^{1 / 5} \\=&[\cos (2 k \pi+3 \alpha)+i \sin (2 k \pi+3 \alpha)]^{1 / 5} \\=&\left[\cos \left(\frac{2 k \pi+3 \alpha}{5}\right)+i \sin \left(\frac{2 k \pi+3 \alpha}{5}\right)\right] \\ & \text { where } k=0,1,2,3,4 \end{aligned}$
Product of all values.
$\begin{array}{l}
=\left(\cos \frac{3 \alpha}{5}+i \sin \frac{3 \alpha}{5}\right) \cdot\left(\cos \left(\frac{2 \pi+3 \alpha}{5}\right)\right. \\
\left.+i \sin \left(\frac{2 \pi+3 \alpha}{5}\right)\right) \cdot\left(\cos \frac{4 \pi+3 \alpha}{5}+i \sin \right. \\
\left.\frac{4 \pi+3 \alpha}{5}\right) \cdot\left(\cos \frac{6 \pi+3 \alpha}{5}+i \sin \frac{6 \pi+3 \alpha}{5}\right) \\
=\cos \left[\frac{3 \alpha}{5}+\frac{2 \pi+3 \alpha}{5}+\frac{4 \pi+3 \alpha}{5}+\frac{6 \pi+3 \alpha}{5}+\right. \\
\left.\frac{8 \pi+3 \alpha}{5}\right]+i \sin \left[\frac{3 \alpha}{5}+\frac{2 \pi+3 \alpha}{5}+\frac{4 \pi+3 \alpha}{5}\right. \\
\left.\quad+\frac{6 \pi+3 \alpha}{5}+\frac{8 \pi+3 \alpha}{5}\right]
\end{array}$
$=\cos \left[\frac{5}{2}\left(2 \cdot \frac{3 \alpha}{5}+\cdot(5-1) \cdot\left(\frac{2 \pi}{5}\right)\right)\right]$
$\quad+i \sin \left[\frac{5}{2}\left(2 \cdot \frac{3 \alpha}{5}+(5-1) \cdot\left(\frac{2 \pi}{5}\right)\right)\right]$
$=\cos \left[\frac{5}{2} \cdot\left(\frac{6 \alpha}{5}+\left(\frac{8 \pi}{5}\right)\right)\right]+$
$=\cos (3 \alpha+4 \pi)+i \sin (3 \alpha+4 \pi)$
$=\cos (4 \pi+3 \alpha)+i \sin (4 \pi+3 \alpha)$
$=\cos 3 \alpha+i \sin 3 \alpha$
Hence, the correct answer is option: 'C'.