If ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}},{\rm{\;where}}\;r = 1,\;2,\;3 \ldots ,\;n,\;\;{\rm{then}}\;\mathop {\lim }\limits_{n \to \infty } {z_1}\;{z_2}\;{z_3} \ldots {z_n}\;$ is equal to

$\cos {\rm{\alpha }} + i\sin {\rm{\alpha }}$

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

$\sqrt[3]{{{e^{i{\rm{\alpha }}}}}}$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Analyzing +4 | -1

{z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}},\mathrm{\;where}\;r = 1,\;2,\;3 \ldots ,\;n,\;\;\mathrm{then}\;\lim\limits_{n \to \infty } {z_1}\;{z_2}\;{z_3} \ldots {z_n}\;

is equal to

View answer and explanation

Correct option: C

{e^{i\mathrm{\alpha }/2}}

Option C is correct. By using the properties of complex numbers and limits, the expression simplifies to

{e^{i{\alpha }/2}}