If $z \in \mathbf{C}$ and $2 z=|z|+i$, then $z$ equals

$\frac{\sqrt{3}}{6}+\frac{1}{3} i$

$\frac{\sqrt{3}}{6}+\frac{1}{6} i$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

z \in \mathbf{C}

and

2 z=|z|+i

, then

z

equals

View answer and explanation

Correct option: A

\frac{\sqrt{3}}{6}+\frac{1}{2} i

Option C (

\frac{{\sqrt{{3}}}}{{6}} + \frac{{1}}{{4}}i

) is incorrect. The correct answer is not among the options provided. The correct solution involves equating real and imaginary parts and solving for z, which results in z = 1/(2

\sqrt{3}

) + i/2.

More Questions from COMPLEX NUMBERS

Practice All Chapter MCQs →

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