Question

If \[z = x + iy\;\] is a complex number satisfying\[\mid z + \dfrac{i}{2}{\mid ^2} = \mid z - \dfrac{i}{2}{\mid ^2}\], then the locus of z is [EAMCET \[2002\]]
A) \[2y = x\]
B) \[y = x\]
C) \[y - axis\]
D) \[x - axis\]

Answer 1

Hint: in this question we have to find what locus of the point in the complex plane represent. First, write the z complex number as a combination of real and imaginary numbers. Put z in form of real and imaginary number into the equation.



Formula Used:Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one



Complete step by step solution:Given: Equation in the form of complex number
Now we have complex number equation\[\mid z + \dfrac{i}{2}{\mid ^2} = \mid z - \dfrac{i}{2}{\mid ^2}\]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number

iy is a imaginary part of complex number
Put this value in\[\mid z + \dfrac{i}{2}{\mid ^2} = \mid z - \dfrac{i}{2}{\mid ^2}\]
\[\mid (x + iy) + \dfrac{i}{2}{\mid ^2} = \mid (x + iy) - \dfrac{i}{2}{\mid ^2}\]
\[\mid x + i(y + \dfrac{1}{2}){\mid ^2} = \mid x + i(y - \dfrac{1}{2}){\mid ^2}\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[{x^2} + {(y + \dfrac{1}{2})^2} = {x^2} + {(y - \dfrac{1}{2})^2}\]
\[2y = 0\]
This equation represents equation of line.
Here \[2y = 0\]represent the equation of x-axis therefore locus of point represent-axis.



Option ‘D’ is correct


Note: Complex number is a number which is a combination of real and imaginary number. So in complex number questions, we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.