JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Hard

1 of 1

Analyzing +4 | -1

The value of the expression

2\left( {1 + \frac{1}{\omega }} \right)\left( {1 + \frac{1}{{{\omega ^2}}}} \right) + 3\left( {2 + \frac{1}{\omega }} \right)\left( {2 + \frac{1}{{{\omega ^2}}}} \right) + \ldots + \left( {n + 1} \right)\left( {n + \frac{1}{\omega }} \right)\left( {n + \frac{1}{{{\omega ^2}}}} \right)

View answer and explanation

Correct option: C

{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + n

Option C (

{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + n

) is correct because by simplifying the expression using properties of complex cube roots of unity, we can derive the given result.

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}