JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

\prod\limits_{p = 1}^r {{e^{ip\theta }} = 1,}

where

\prod

denotes the continued product and

i=\sqrt{-1}

, the most general value of

\theta

is
(where,

n

is an integer)

View answer and explanation

Correct option: D

\frac{4 n \pi}{r(r+1)}, n \in I

Option D (

\frac{{4 n \pi }}{{r(r + 1)}}, n \in I

) is correct because it correctly sums the geometric series in the exponent and solves for

\theta

using the properties of complex exponentials.

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