Question

If ${\Pi }_{p=1}^r{e}^{ip\theta }=1$ where $\Pi$ denotes the continued product, then the most general value of $\theta$ is

• A. $\frac{2n\pi }{r(r-1)}$
• B. $\frac{2n\pi }{r(r+1)}$
• C. $\frac{4n\pi }{r(r-1)}$
• D. $\frac{4n\pi }{r(r+1)}$

Answer 1

The correct option is D $\frac{4n\pi }{r(r+1)}$

$\prod _{p=1}^r{e}^{ip\theta }=1$
$\Rightarrow \prod _{p=1}^r{e}^{ip\theta }=\cos 2n\pi +i\sin 2n\pi$
$\Rightarrow e^{i\theta {\sum }_{p=1}^{r}p}=e^{i2n\pi }$
$\Rightarrow e^{\frac{i\theta r(r+1)}{2}}=e^{i2n\pi }$
$\Rightarrow \frac{r(r+1)\theta }{2}=2n\pi$
$\therefore \theta =\frac{4n\pi }{r(r+1)}$
Hence, option 'D' is correct.