Question

If z and w be two complex numbers, such that $|z{|^2}w - |w{|^2}z = z - w{\text{ and, (}}z \ne w{\text{)}}$ then
$A. z = \bar w \\\ B. z\bar w = 1 \\\ C. z\bar zw = 2 \\\ D. z\bar w = 4 \\\$

Answer 1

A complex number is a number generally represented as$z = a + ib$, where $a$ and $b$ is a real number represented on the real axis whereas $i$ is an imaginary unit represented on the imaginary axis whose value is $i = \sqrt { - 1}$. Modulus of a complex number is the length of a line segment on a real and imaginary axis generally denoted by $|z| = \sqrt {{a^2} + {b^2}}$. Conjugate of a complex number is the negation of the imaginary part of the complex number while keeping the real part as it is such as for $z = a + ib$, the complex conjugate will be defined as $z = a - ib$.
In this question, two complex numbers are given, and we need to determine the relation between the conjugate of one complex number with the original form of another complex number for which we need to follow the concept of argument and properties of the complex conjugate of the numbers.

Complete step by step solution: Rearranging the given equation $|z{|^2}w - |w{|^2}z = z - w{\text{ and, (}}z \ne w{\text{)}}$ as:

$$|z{|^2}w - |w{|^2}z = z - w{\text{ }} \\\ |z{|^2}w + w = z + |w{|^2}z \\\ w\left( {|z{|^2} + 1} \right) = z\left( {|w{|^2} + 1} \right) \\\ \dfrac{w}{z} = \left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right) - - - - (i) \\\$$

As equation (i), does not have any imaginary part so, from equation (i), it can also be written as:
$\dfrac{{\bar w}}{{\bar z}} = \left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right) - - - - (ii)$
Dividing equation (i) and equation (ii) we get,

$$\dfrac{{\left( {\dfrac{w}{z}} \right)}}{{\left( {\dfrac{{\bar w}}{{\bar z}}} \right)}} = \dfrac{{\left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right)}}{{\left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right)}} \\\ \dfrac{{w\bar z}}{{\bar wz}} = 1 \\\ w\bar z = \bar wz = 1 \\\$$

Hence, for two complex numbers $z{\text{ and }}w$, such that $|z{|^2}w - |w{|^2}z = z - w{\text{ and, (}}z \ne w{\text{)}}$ then $z\bar w = 1$.

Option B is correct.

Note: It is interesting to note here that, the complex conjugate of a real number is the number itself as it does not contain any imaginary term with it such as for $z = pq$, the complex conjugate will be the number itself i.e., $z = pq$ as it does not have any imaginary part for the negation.