Differentiation Question 192
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{x\to \infty }{\mathop{\lim }}z_1z_2z_3…z _{n} $ is equal to
Options:
A) $ \cos \alpha +i\sin \alpha $
B) $ \cos (\alpha /2)-isin(\alpha /2) $
C) $ {e^{i\alpha /2}} $
D) $ \sqrt[3]{{e^{i\alpha }}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ z _{r}=\cos \frac{ra}{n^{2}}+i\sin \frac{ra}{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}};…, $
$ z _{n}=\cos \frac{n\alpha }{n^{2}}+i\sin \frac{n\alpha }{n^{2}} $ Consider $ \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{an(n+1)}{2n^{2}}+i\sin \frac{an(n+1)}{2n^{2}} ] $
$ =\underset{x\to \infty }{\mathop{\lim }}\frac{\cos a( 1+1/n )}{2}+\frac{i\sin a( 1+1/n )}{2} $
$ =\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2}={e^{\frac{ia}{2}}} $
BETA
