JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

Let

\alpha

and

\beta

be two fixed non-zero complex numbers and '

z

' a variable complex number. If the lines

\alpha \bar z + \bar \alpha z + 1 = 0

and

\mathrm{\beta \;\bar z} + \bar \beta \;z - 1 = 0

are mutually perpendicular, then

View answer and explanation

Correct option: D

\alpha \;\bar \beta + \bar \alpha \;\mathrm{\beta } = 0

Option D (

\alpha \;\bar \beta + \bar \alpha \;\mathrm{\beta } = 0

) is correct because it represents the condition for the lines to be perpendicular, which involves the sum of the product of the coefficients and their conjugates being 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