JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

{z_r} = \cos \left( {\frac{\pi }{{{2^r}}}} \right) + i\sin \left( {\frac{\pi }{{{2^r}}}} \right)

, then

{z_1}.{z_2}.{z_3} \ldots

upto

\infty

equals

View answer and explanation

Correct option: D

- 1

Option D ( - 1) is correct because the infinite product of the given complex numbers simplifies to -1 using trigonometric identities.

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

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\alpha, \beta

\gamma

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