Question

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}} $ upto $ \infty $ equals:

• A. $ -3 $
• B. $ -2 $
• C. 1
• D. 0
• E. -1

Answer 1

Answer: (E)

$ {{z}_{1}}.{{z}_{2}}.{{z}_{3}}....\infty $
$ =\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\left( \cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}} \right) $ $ \times \left( \cos \frac{\pi }{{{2}^{3}}}+i\sin \frac{\pi }{{{2}^{3}}} \right)... $
$ =\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{\pi /2}{1-1/2} \right)+i\sin \left( \frac{\pi /2}{1-1/2} \right) $ $ =\cos \pi +i\sin \pi =-1 $