JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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{z_1},\;{z_2},{z_3},\;{z_4}

are the affixes of four points in the argand plane and

z

is the affix of a point such that

\left| {\;z - {z_1}} \right| = \left| {z - {z_2}} \right| = \left| {z - {z_3}} \right| = \left| {z - {z_4}} \right|,\;\mathrm{then}\,\;{z_1},{z_2},\;{z_3},{z_4}

are

View answer and explanation

Correct option: A

Concyclic

Option B is correct because Rolle's Theorem guarantees the existence of a point 'c' between two roots 'a' and 'b' where the derivative of the function is zero.

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