JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Analyzing +4 | -1

\omega

is complex cube root of unity and

a, b, c

are such that

\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2 \omega^2

and

\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2 \omega

, then the value of

\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}

is equal to

View answer and explanation

Correct option: D

2

Option D (2) is correct because by recognizing omega and omega squared as roots of an equation, the third root can be found and used to evaluate the expression.

More Questions from COMPLEX NUMBERS

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The number of non-zero integer solutions of the equation

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

is If

|z|=1

and

z \neq \pm 1

, then for any complex number

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

Let

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

, then

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

is less than The roots of the cubic equation

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

, where

a \neq 0

represents the vertices of a triangle. If

\alpha, \beta

and

\gamma

are the roots, then find the value of

|\alpha-\beta|

. Number of solutions of the equation

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

where

z \in C

is If

2x = - 1 + \sqrt 3 i

, then the value of

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

is equal to