Complex Numbers And Quadratic Equations question 156
Question: If $ z_{r}=\cos \frac{r\alpha }{n^{2}}+i\sin \frac{r\alpha }{n^{2}}, $ where r = 1, 2, 3,……,n, then $ \underset{n\to \infty }{\mathop{\lim }}z_1z_2z_3…z_{n} $ is equal to [UPSEAT 2001]
Options:
A) $ \cos \alpha +i\sin \alpha $
B) $ \cos (\alpha /2)-i\sin (\alpha /2) $
C) $ {e^{i\alpha /2}} $
D) $ \sqrt[3]{{e^{i\alpha }}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ z_{r}=\cos \frac{r\alpha }{n^{2}}+i\sin \frac{r\alpha }{n^{2}} $ $ z_1=\cos \frac{\alpha }{n^{2}}+i\sin \frac{\alpha }{n^{2}} $ ; $ z_2=\cos \frac{2\alpha }{n^{2}}+i\sin \frac{2\alpha }{n^{2}} $ ; ….
Þ $ z_{n}=\cos \frac{n\alpha }{n^{2}}+i\sin \frac{n\alpha }{n^{2}} $
$ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}(z_1z_2z_3………z_{n}) $ $ =\underset{n\to \infty }{\mathop{\lim }}[ \cos { \frac{\alpha }{n^{2}}(1+2+3+…+n) } . $ $ . +i\sin { \frac{\alpha }{n^{2}}(1+2+3+…+n) } ] $ $ =\underset{n\to \infty }{\mathop{\lim }}[ \cos \frac{\alpha n(n+1)}{2n^{2}}+i\sin \frac{\alpha n(n+1)}{2n^{2}} ] $ $ =\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2}={e^{\frac{i\alpha }{2}}} $ .
BETA
