Question

A complex number z is said to be unimodular if $\left| z \right|=1$ . Suppose ${{z}_{1}}$ and ${{z}_{2}}$ are complex numbers such that $\dfrac{{{z}_{1}}-2{{z}_{2}}}{2-{{z}_{1}}{{{\bar{z}}}_{2}}}$is unimodular and ${{z}_{2}}$ is not unimodular, then the point ${{z}_{1}}$lies on a :
( a ) straight line parallel to y – axis
( b ) circle of radius 2.
( c ) circle of radius $\sqrt{2}$
( d ) straight line parallel to x - axis

Answer 1

What we will do in this question is, we will use properties of complex number such as $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$, $z\bar{z}={{\left| z \right|}^{2}}$and property of unimodular which is $\left| z \right|=1$to solve the question and break down the complex equation into a simpler form of equation.

Complete step-by-step answer:
Now, if z is unimodular then $\left| z \right|=1$ and $z\bar{z}={{\left| z \right|}^{2}}$.
Now, in question it is given that ${{z}_{2}}$is not unimodular, so $\left| {{z}_{2}} \right|\ne 1$and also it is given that $\dfrac{{{z}_{1}}-2{{z}_{2}}}{2-{{z}_{1}}{{{\bar{z}}}_{2}}}$is unimodular.
So, $\left| \dfrac{{{z}_{1}}-2{{z}_{2}}}{2-{{z}_{1}}{{{\bar{z}}}_{2}}} \right|=1$
By cross multiplication, we get
$\left| {{z}_{1}}-2{{z}_{2}} \right|=\left| 2-{{z}_{1}}{{{\bar{z}}}_{2}} \right|$
Squaring both side, we get
${{\left| {{z}_{1}}-2{{z}_{2}} \right|}^{2}}={{\left| 2-{{z}_{1}}{{{\bar{z}}}_{2}} \right|}^{2}}$
Using property, $z\bar{z}={{\left| z \right|}^{2}}$, we get
$\left( {{z}_{1}}-2{{z}_{2}} \right)\left( {{{\bar{z}}}_{1}}-2{{{\bar{z}}}_{2}} \right)=\left( 2-{{z}_{1}}{{{\bar{z}}}_{2}} \right)(2-{{z}_{2}}{{\bar{z}}_{1}})$
On opening and solving brackets, we get
${{z}_{1}}{{\bar{z}}_{1}}-2{{z}_{1}}{{\bar{z}}_{2}}-2{{z}_{2}}{{\bar{z}}_{1}}+4{{z}_{2}}{{\bar{z}}_{2}}=4-2{{z}_{2}}{{\bar{z}}_{1}}-2{{z}_{1}}{{\bar{z}}_{2}}+{{z}_{1}}{{\bar{z}}_{1}}\cdot {{z}_{2}}{{\bar{z}}_{2}}$
On simplifying, we get
${{z}_{1}}{{\bar{z}}_{1}}+4{{z}_{2}}{{\bar{z}}_{2}}=4+{{z}_{1}}{{\bar{z}}_{1}}\cdot {{z}_{2}}{{\bar{z}}_{2}}$
Using, property, $z\bar{z}={{\left| z \right|}^{2}}$, we get
${{\left| {{z}_{1}} \right|}^{2}}+4{{\left| {{z}_{2}} \right|}^{2}}=4+{{\left| {{z}_{1}} \right|}^{2}}\cdot {{\left| {{z}_{2}} \right|}^{2}}$
Re – writing above equation, we get
${{\left| {{z}_{1}} \right|}^{2}}\cdot {{\left| {{z}_{2}} \right|}^{2}}+4-{{\left| {{z}_{1}} \right|}^{2}}-4{{\left| {{z}_{2}} \right|}^{2}}=0$
Again,
${{\left| {{z}_{1}} \right|}^{2}}\cdot {{\left| {{z}_{2}} \right|}^{2}}+4-{{\left| {{z}_{1}} \right|}^{2}}-4{{\left| {{z}_{2}} \right|}^{2}}=0$
${{\left| {{z}_{1}} \right|}^{2}}\cdot ({{\left| {{z}_{2}} \right|}^{2}}-1)-4({{\left| {{z}_{2}} \right|}^{2}}-1)=0$
$({{\left| {{z}_{1}} \right|}^{2}}-4)\cdot ({{\left| {{z}_{2}} \right|}^{2}}-1)=0$
As, in question it is given that ${{z}_{2}}$is not unimodular, so $\left| {{z}_{2}} \right|\ne 1$
So, $({{\left| {{z}_{2}} \right|}^{2}}-1)\ne 0$
Thus, $({{\left| {{z}_{1}} \right|}^{2}}-4)=0$
${{\left| {{z}_{1}} \right|}^{2}}=4$
Or, $\left| {{z}_{1}} \right|=2$
We know that , $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
So, $\left| {{z}_{1}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=2$
$\sqrt{{{x}^{2}}+{{y}^{2}}}=2$
${{x}^{2}}+{{y}^{2}}={{2}^{2}}$, which is equation of a circle whose centre is ( 0, 0 ) and radius is 2 as general equation of circle whose center is ( h, k ) and radius is r is ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

So, the correct answer is “Option b”.

Note: These proving types of questions based on complex numbers are very tricky and lengthy so always put remembering formula and basic theory of complex numbers as priority. Always remember that if z = x + iy then $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$, $z\bar{z}={{\left| z \right|}^{2}}$ and general equation of circle whose centre is ( h, k ) and radius is r is ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$ . Try to avoid calculation mistakes else the solution will get wrong.