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Question: A complex number z is said to be unimodular if |z| \( \ne 1 \) . If \( {z_1}\,\,and\,\,{z_2} \) are ...

A complex number z is said to be unimodular if |z| ≠1\ne 1 . If z1  and  z2{z_1}\,\,and\,\,{z_2} are complex numbers such that z1−2z22−z1z‾2\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}{{\overline z }_2}}} is unimodular and z2{z_2} is not unimodular.
Then, the points z1{z_1} lies on a.
(a) Straight line parallel to X-axis
(b) Straight line parallel to Y-axis
(c) Circle of radius 22
(d) Circle of radius 2\sqrt 2

Explanation

Solution

Hint : Since, it is given that z1−2z22−z1z‾2\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}{{\overline z }_2}}} is unimodular. Therefore equate its mode equal to one and then remove mode by doing squaring both sides and on simplifying by using the following mentioned formulas of complex one can easily find the correct option.
For  any  z=x+iy,    ∣z∣2=z.z‾,      and  ∣x+iy∣=x2+y2For\,\,any\,\,z = x + iy,\,\,\,\,|z{|^2} = z.\overline z ,\,\,\,\,\,\,and\,\,|x + iy| = \sqrt {{x^2} + {y^2}}

Complete step-by-step answer :
Since it is given that z1−2z22−z1z‾2\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}{{\overline z }_2}}} is unimodular.
Therefore z1−2z22−z1z‾2=1 {\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}{{\overline z }_2}}}} = 1
⇒∣z1−2z2∣=∣2−z1z‾2∣\Rightarrow |{z_1} - 2{z_2}| = |2 - {z_1}{\overline z _2}|
To remove mode from the above equation. We do squaring both side

(z1−2z2)2=(2−z1z‾2)2 ⇒(z1−2z2).(z‾1−2z‾2)=(2−z1z‾2).(2−z‾1z2) ⇒z1z‾1−2.z2.z‾1−2.z1.z‾2+4.z2.z‾2=4−2.z2.z‾1−2.z1.z‾2+z‾1.z2z1z‾2 ⇒(z1)2+4.(z2)2=4+z‾1.z2z1z‾2 ⇒∣z1∣2+4∣z2∣2=4+∣z1∣2∣z2∣2 ⇒∣z1∣2+4∣z2∣2−4−∣z1∣2∣z2∣2=0 ⇒∣z1∣2−∣z1∣2∣z2∣2+4∣z2∣2− 4=0 ⇒∣z1∣2(1−∣z2∣2)−4(1−∣z2∣2)=0 ⇒(1−∣z2∣2)(∣z1∣2−4)=0 ⇒(1−∣z2∣2)=0    or  (∣z1∣2−4)=0 ⇒∣z2∣2=1  or  ∣z1∣2=4 ⇒∣z2∣=1   or  ∣z1∣=2  {\left( {{z_1} - 2{z_2}} \right)^2} = {\left( {2 - {z_1}{{\overline z }_2}} \right)^2} \\\ \Rightarrow \left( {{z_1} - 2{z_2}} \right).\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right) = \left( {2 - {z_1}{{\overline z }_2}} \right).\left( {2 - {{\overline z }_1}{z_2}} \right) \\\ \Rightarrow {z_1}{\overline z _1} - 2.{z_2}.{\overline z _1} - 2.{z_1}.{\overline z _2} + 4.{z_2}.{\overline z _2} = 4 - 2.{z_2}.{\overline z _1} - 2.{z_1}.{\overline z _2} + {\overline z _1}.{z_2}{z_1}{\overline z _2} \\\ \Rightarrow {({z_1})^2} + 4.{({z_2})^2} = 4 + {\overline z _1}.{z_2}{z_1}{\overline z _2} \\\ \Rightarrow |{z_1}{|^2} + 4|{z_2}{|^2} = 4 + |{z_1}{|^2}|{z_2}{|^2} \\\ \Rightarrow |{z_1}{|^2} + 4|{z_2}{|^2} - 4 - |{z_1}{|^2}|{z_2}{|^2} = 0 \\\ \Rightarrow |{z_1}{|^2} - |{z_1}{|^2}|{z_2}{|^2} + 4|{z_2}{|^2} - \,4 = 0 \\\ \Rightarrow |{z_1}{|^2}\left( {1 - |{z_2}{|^2}} \right) - 4\left( {1 - |{z_2}{|^2}} \right) = 0 \\\ \Rightarrow \left( {1 - |{z_2}{|^2}} \right)\left( {|{z_1}{|^2} - 4} \right) = 0 \\\ \Rightarrow \left( {1 - |{z_2}{|^2}} \right) = 0\,\,\,\,or\,\,\left( {|{z_1}{|^2} - 4} \right) = 0 \\\ \Rightarrow |{z_2}{|^2} = 1\,\,or\,\,|{z_1}{|^2} = 4 \\\ \Rightarrow |{z_2}| = 1\,\,\,or\,\,|{z_1}| = 2 \\\

But since it is given that z2≠1{z_2} \ne 1
Therefore, we have ∣z1∣=2|{z_1}| = 2
Or we can write it as (x2+y2)=2            (∵∣z∣=x2+y2)\sqrt {({x^2} + {y^2})} = 2\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because |z| = \sqrt {{x^2} + {y^2}} } \right)
Squaring both side we have
x2+y2=4{x^2} + {y^2} = 4
Which is an equation of circle having centre at origin and of radius 2 units2\,units .
Hence, from above we see the out of the given four options, option (C) is the correct option.

So, the correct answer is “Option C”.

Note : We know that for complex numbers we always take z = x + iy and to find solutions to any complex problems we start with the given condition and then simplify the given problem by using different properties of complex numbers.