If $z$ is a complex number satisfying $\left| {z - i\operatorname{Re} \left( z \right)} \right| = \left| {z - \operatorname{Im} \left( z \right)} \right|$ then $z$ lies on
A. $y = x$
B. $y = - x$
C. $y = x + 1$
D. $y = - x + 1$

Hint: The complex number is written in the form of $a + ib$, where $a$ and $b$ are real numbers and $i$ is iota. In $z = a + ib$, $a$ is called the real part and $b$ is called the imaginary part. Let us first assume that $z = x + iy$ be a complex number. Then, the modulus of $z$, is denoted by $\left| z \right|$, is defined to be the non-negative real number $\sqrt {{x^2} + {y^2}} $, i.e., $\left| z \right| = \sqrt {{x^2} + {y^2}} $. We will substitute the value of $z = x + iy$ in the given expression and simplify it. After this, we will apply $\left| z \right| = \sqrt {{x^2} + {y^2}} $ formula to get our required answer.

Complete step by step solution:
 We have, $\left| {z - i\operatorname{Re} \left( z \right)} \right| = \left| {z - \operatorname{Im} \left( z \right)} \right|$
Let the complex number be $z = x + iy$.
$ \Rightarrow \left| {x + iy - i\operatorname{Re} \left( {x + iy} \right)} \right| = \left| {x + iy - \operatorname{Im} \left( {x + iy} \right)} \right|$
We know that the real part of $x + iy$ is $x$ and the imaginary part of $x + iy$ is $y$. Therefore, we get
$ \Rightarrow \left| {x + iy - ix} \right| = \left| {x + iy - y} \right|$
We can write the above written expression as
$ \Rightarrow \left| {x + i\left( {y - x} \right)} \right| = \left| {\left( {x - y} \right) + iy} \right|$
We know that $\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} $. Therefore, we get
$ \Rightarrow \sqrt {{x^2} + {{\left( {y - x} \right)}^2}} = \sqrt {{{\left( {x - y} \right)}^2} + {y^2}} $
On squaring both the sides, we get
$ \Rightarrow {x^2} + {\left( {y - x} \right)^2} = {\left( {x - y} \right)^2} + {y^2}$
We know that ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. Therefore, we get
$ \Rightarrow {x^2} + {y^2} + {x^2} - 2yx = {x^2} + {y^2} - 2yx + {y^2}$
On canceling the common terms, we get
$ \Rightarrow {x^2} = {y^2}$
$\therefore x = y$
Hence,$z$ lies on $x = y$.

Option ‘A’ is correct

Note: As in the above question $i$ i.e., iota is given, we can’t put the value of $i$ here because the value of $i$ is $\sqrt { - 1} $. If we substitute the value of $i$ here, it will make the question more complex. Here, modulus plays the most important role. Don’t just remove the modulus, solve the modulus by using formula $\left| z \right| = \sqrt {{x^2} + {y^2}} $, where $z = x + iy$. Take care of the calculation, so as to be sure of your final answer.