JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Medium

1 of 1

Application +4 | -1

\left(a_1+i b_1\right)\left(a_2+i b_2\right) \ldots\left(a_n+i b_n\right)=A+i B

, then

\sum\limits_{{j=1}}^n \tan ^{-1}\left(\frac{b_j}{a_j}\right)

is equal to

View answer and explanation

Correct option: C

\tan ^{-1}\left(\frac{B}{A}\right)

Option C (

\tan ^{-1}\left(\frac{{B}}{{A}}\right)

) is correct because the argument of a product of complex numbers is the sum of their arguments.

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}