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JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

If

f(x)=a x^2+b x

, where

f(1)=2, f\left(i^2\right)=0

and

f\left(z+z^{-1}\right)=0

, then

|z|=

View answer and explanation

Correct option: D

1

Option D (1) is correct because solving the equations

f(1)=2

and

f(i^2)=0

gives

f(x) = x^2 + x

, and solving

f(z+z^{-1})=0

leads to

|z|=1

.

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The number of non-zero integer solutions of the equation

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

|z|=1

z \neq \pm 1

, then for any complex number

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|

is less than The roots of the cubic equation

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

a \neq 0

represents the vertices of a triangle. If

\alpha, \beta

\gamma

are the roots, then find the value of

|\alpha-\beta|

. Number of solutions of the equation

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

z \in C

2x = - 1 + \sqrt 3 i

, then the value of

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