Question

The complex number $z = x + iy$ which satisfies the equation $\left| {\dfrac{{z - 3i}}{{z + 3i}}} \right| = 1$ lies on
A. The $x -$ axis
B. The straight line $y = 3$
C. A circle passing through origin
D. None of the above

Answer 1

In the above question we have been given
$\left| {\dfrac{{z - 3i}}{{z + 3i}}} \right| = 1$ . So to solve this question we will use the property of modulus of complex numbers and use given conditions to create equations to find the solution.
We will use the property:
$\left| {x + iy} \right| = \sqrt {{x^2} + {{(iy)}^2}}$ .

Complete step-by-step solution:
Let us first understand the definition of complex numbers.
We know that a complex number is a number that can be expressed in the form of
$a + ib$ , where $a,b$ are real numbers and $i$ is the symbol of an imaginary unit.
So here we have $\left| {\dfrac{{z - 3i}}{{z + 3i}}} \right| = 1$
We can write the above expression also as
$\left| {z - 3i} \right| = \left| {z + 3i} \right|$
We will now put the value of
$z = x + iy$ in the above equation and we have:
$\left| {x + iy - 3i} \right| = \left| {x + iy + 3i} \right|$
We can separate the real part and the imaginary part i.e.
$\left| {x + i(y - 3)} \right| = \left| {x + i(y + 3)} \right|$
By using the above property or by squaring both the sides we can write:
$\sqrt {{x^2} + {{\left( {i(y - 3)} \right)}^2}} = \sqrt {{x^2} + {{\left( {i(y + 3)} \right)}^2}}$
We should note that the value of $i$ is
$\sqrt { - 1}$ .
This is the imaginary number, also called iota. So if we square the value of iota i.e.
${i^2}$ , we get the value
$\sqrt { - 1} \times \sqrt { - 1} = - 1$ .
So it gives us
${i^2} = - 1$ .
By putting this back in the expression we can write
$\sqrt {{x^2} - {{\left( {y - 3} \right)}^2}} = \sqrt {{x^2} - {{\left( {y + 3} \right)}^2}}$
Now by squaring both the sides, we get
${x^2} - {\left( {y - 3} \right)^2} = {x^2} - {\left( {y + 3} \right)^2}$
We can eliminate same terms from both the sides of the above equations, and we have:
${\left( {y - 3} \right)^2} = {\left( {y + 3} \right)^2}$
By breaking the values we have:
${x^2} + {y^2} + 9 - 6y = {x^2} + {y^2} + 9 + 6y$
By eliminating the similar terms again, we can write
$6y + 6y = 0$
It gives us
$12y = 0 \Rightarrow y = \dfrac{0}{{12}} = 0$
Now we know that the value
$y = 0$ represents the x- axis.
Hence the correct option is (a) The $x -$ axis.

Note: We should note that in the above solution we have used the algebraic formula i.e.
${(a + b)^2} = {a^2} + {b^2} + 2ab$ and
${(a - b)^2} = {a^2} + {b^2} - 2ab$
There is also an alternate way to solve the final part of the solution. We can write ${\left( {y - 3} \right)^2} = {\left( {y + 3} \right)^2}$ as ${\left( {y - 3} \right)^2} - {\left( {y + 3} \right)^2} = 0$
Here we will use the property
${a^2} - {b^2} = (a - b)(a + b)$
So by putting this in the above equation:
$(y - 3 - y - 3)(y - 3 + y + 3) = 0$ .
It will give us
$- 6y = 0$
Or,
$y = \dfrac{0}{{ - 6}} = 0$ .
This way also we have $x -$ axis.