JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Analyzing +4 | -1

Let

a = {e^{i\frac{{2\pi }}{3}\;}}

Then, the equation whose roots are

a + {a^{ - 2}}

and

{a^2} + {a^{ - 4}}

View answer and explanation

Correct option: D

{x^2} + 2x + 4 = 0

Option D (

{x^2} + 2x + 4 = 0

) is the correct equation, found by determining the roots

a + a^{-2}

and

a^2 + a^{-4}

and then constructing the quadratic equation.

More Questions from COMPLEX NUMBERS

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

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z \neq \pm 1

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\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

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\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}