If $z_{r} = \cos{\left( \frac{\pi}{2^{r}} \right) + i\sin\left( \frac{\pi}{2^{r}} \right)}$, then $z_{1}.z_{2}.z_{3}\ldots$ upto $\infty$ equals

Question

Dr. Abubacker · Accepted Answer

(d)
$z_{1}.z_{2}.z_{3},\ \ldots\ \infty$
$= \cos\left( \frac{\pi}{2} + \frac{\pi}{2^{2}} + \frac{\pi}{2^{3}} + ... \right) + i\sin\left( \frac{\pi}{2} + \frac{\pi}{2^{2}} + \frac{\pi}{2^{3}} + ... \right)$
$= \cos\left( \frac{\frac{\pi}{2}}{1 - \frac{1}{2}} \right) + i\sin\left( \frac{\frac{\pi}{2}}{1 - \frac{1}{2}} \right)$
$= \cos{\pi + i\sin{\pi = - 1}}$

Q. If $z_{r} = \cos{\left( \frac{\pi}{2^{r}} \right) + i\sin\left( \frac{\pi}{2^{r}} \right)}$, then $z_{1}.z_{2}.z_{3}\ldots$ upto $\infty$ equals

Explanation

Reviewed by Dr. Abubacker

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