If ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}}$, where $r = 1,2,3, \ldots ,n$, then $\mathop {\lim }\limits_{n \to \infty } {z_1}{z_2} \ldots {z_n}$ is equal to

$\cos \left( {\frac{\alpha }{2}} \right) - i\sin \left( {\frac{\alpha }{2}} \right)$

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

{z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}}

, where

r = 1,2,3, \ldots ,n

, then

\lim\limits_{n \to \infty } {z_1}{z_2} \ldots {z_n}

is equal to

View answer and explanation

Correct option: C

{e^{i\alpha /2}}

Option C (

{e^{i\alpha /2}}

) is correct because by using Euler's formula and simplifying the sum, the limit converges to

{e^{i\alpha /2}}