Question

The locus of the point $z = x + iy$ satisfying the equation $\left| {\dfrac{{z - 1}}{{z + 1}}} \right| = 1$ is given by
A.$x = 0$
B.$y = 0$
C.$x = y$
D.$x + y = 0$

Answer 1

Hint : In this equation we have to find the locus of the point$z = x + iy$which satisfies the equation$\left| {\dfrac{{z - 1}}{{z + 1}}} \right| = 1$, so first we will substitute the point in the given equation and then we will find its modulus and then by further solving the obtained equation we will find the locus of the given point.

Complete step-by-step answer :
Given the equation $\left| {\dfrac{{z - 1}}{{z + 1}}} \right| = 1$
The point $z = x + iy$, where x is the real part and the y is an imaginary number.
Now we substitute the locus point in the given equation, we get
$\left| {\dfrac{{\left( {x + iy} \right) - 1}}{{\left( {x - iy} \right) + 1}}} \right| = 1$
Hence by solving this equation, we get
$\left| {\dfrac{{\left( {x - 1} \right) + iy}}{{\left( {x + 1} \right) - iy}}} \right| = 1 - - (i)$
Now by cross multiplying the obtained equation we can further write it as
$\left| {\left( {x - 1} \right) + iy} \right| = 1\left| {\left( {x + 1} \right) - iy} \right| - - (ii)$
Now since we know the modulus of $\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}}$, hence in the same way we can write equation (ii) which is in modulus form as
$\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} = \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( { - y} \right)}^2}}$
Hence by further solving this obtained equation we get

$$\Rightarrow {\left( {x - 1} \right)^2} + {y^2} = {\left( {x + 1} \right)^2} + {\left( { - y} \right)^2} \\\ \Rightarrow {x^2} - 2x + 1 + {y^2} = {x^2} + 2x + 1 + {y^2}\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}} \right] \;$$

By further solving this equation

$$\- 2x = 2x \\\ 0 = 2x + 2x \\\ 4x = 0 \\\ x = 0 \;$$

Hence by substituting this value in the locus point$z = x + iy$, we get
$z = y$
So it lie on the y–axis, hence the locus of the point satisfying the point $z = x + iy$ is x=0

So, the correct answer is “Option A”.

Note : Locus is the set of all points that satisfies a certain condition so here we need to find the point which satisfies the given equation. Students must note that in a complex number$z = x + iy$, if the imaginary part is $y = 0$then locus will lie on real axis and if $x = 0$ then it will lie on imaginary axis
$\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}}$
$\left| {x - iy} \right| = \sqrt {{x^2} + {{\left( { - y} \right)}^2}} = \sqrt {{x^2} + {y^2}}$