JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Medium

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 \rightarrow \infty}}\left(z_1 z_2 z_3 \ldots z_n\right)

is equal to

View answer and explanation

Correct option: C

e^{i a / 2}

Option C (

e^{i a / 2}

) is correct because the product of complex numbers in polar form simplifies to

e^{i \alpha / 2}

after applying trigonometric identities and limit properties.

More Questions from COMPLEX NUMBERS

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}