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Question: State whether the following statement is true or false: Let z\(_1\) and z\(_2\) be two complex num...

State whether the following statement is true or false:
Let z1_1 and z2_2 be two complex numbers such that z12z22z1z2\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }} is unimodular. If z2_2 is not unimodular then z1=2|{z_1}| = 2.

Explanation

Solution

We will first write the definition of unimodular. Then we will use the fact that z.zˉ=z2z.\bar z = |z{|^2}. Then apply the conditions of unimodular on the given expression z12z22z1z2\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }} and finally find the value of z1_1.

Complete step-by-step answer:
Let us first get to know about what is zz and zˉ\bar z.
zz is a complex number which has a real part and an imaginary part. For example:- say z=x+iyz = x + iy, then xx is the real part and yy is the imaginary part.
zˉ\bar z is known to be conjugate of zz, which means that its real part remains as same as of the zz, but the imaginary part becomes negative. So, zˉ=xiy\bar z = x - iy.
Let us now use the fact that z.zˉ=z2z.\bar z = |z{|^2}……(1)
We are already given that z2_2 is not unimodular. So, |z2_2|2^2 \ne 1. …..(2)
Now let us decipher the definition of unimodular.
Unimodular: A complex number is said to be unimodular if its modulus is equal to 1 that if a complex number zz is unimodular, then z=1|z| = 1.
Since, we are given that z12z22z1z2\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }} is unimodular.
So, z12z22z1zˉ2=1\dfrac{{|{z_1} - 2{z_2}|}}{{|2 - {z_1}{{\bar z}_2}|}} = 1.
z12z2=2z1zˉ2\therefore |{z_1} - 2{z_2}| = |2 - {z_1}{\bar z_2}|
Squaring both sides, we will get:-
z12z22=2z1zˉ22|{z_1} - 2{z_2}{|^2} = |2 - {z_1}{\bar z_2}{|^2}
Now using (1), we have:-
(z12z2)(zˉ12zˉ2)=(2z1zˉ2)(2zˉ1z2)({z_1} - 2{z_2})({\bar z_1} - 2{\bar z_2}) = (2 - {z_1}{\bar z_2})(2 - {\bar z_1}{z_2})
Simplifying both the sides, we will get:-
z1zˉ12zˉ1z22z1zˉ2+4z2zˉ2=42z1zˉ22zˉ1z2+z1zˉ1z2zˉ2{z_1}{\bar z_1} - 2{\bar z_1}{z_2} - 2{z_1}{\bar z_2} + 4{z_2}{\bar z_2} = 4 - 2{z_1}{\bar z_2} - 2{\bar z_1}{z_2} + {z_1}{\bar z_1}{z_2}{\bar z_2}
Now using (1) and cancelling out the common terms from both the sides:-
z12+4z22=4z12z22|{z_1}{|^2} + 4|{z_2}{|^2} = 4 - |{z_1}{|^2}|{z_2}{|^2}
z12+4z22+z12z224=0|{z_1}{|^2} + 4|{z_2}{|^2} + |{z_1}{|^2}|{z_2}{|^2} - 4 = 0
(z221)+(z124)=0(|{z_2}{|^2} - 1) + (|{z_1}{|^2} - 4) = 0
So, either (z221)=0(|{z_2}{|^2} - 1) = 0 or (z124)=0(|{z_1}{|^2} - 4) = 0
Using (2), we see that the only possibility is:
(z124)=0(|{z_1}{|^2} - 4) = 0 …….(3)
Let z1=x+iy{z_1} = x + iy
Since, z1=x2+y2|{z_1}| = \sqrt {{x^2} + {y^2}}
Hence, z12=(x2+y2)2=x2+y2|{z_1}{|^2} = {\left( {\sqrt {{x^2} + {y^2}} } \right)^2} = {x^2} + {y^2}
Putting this in (3), we will get:-
(x2+y2{x^2} + {y^2} - 4) = 0
So, x2+y2{x^2} + {y^2} = 4.
This equation depicts a circle with center (0, 0) and radius 2.
Because the general equation of a circle is given by (xh)2+(yk)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}, where (h,k)(h,k) is the center of the circle and rr is the radius.
Hence, z1{z_1} lies on a circle and z1{z_1} = 2.
Hence, the statement given is true.

Note: Remember the definition of Unimodular
Also remember the fact that z.zˉ=z2z.\bar z = |z{|^2}.
There is a possibility that sometimes, we may get some equation of circle with center being non origin, then we have to carefully examine the equation, by completing the square method and find the value of radius.
We might make the mistake of considering both the cases in equation (3),but we were already given the data to reject one of the conditions. We must take care of that.
We must read the question properly as there might be cases that we are already given no in the question. For example:- State whether the following statement is true or false:
Let z1_1 and z2_2 be two complex numbers such that z12z22z1z2\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }} is unimodular. If z2_2 is not unimodular then z1|{z_1}| \ne 2.
In this case our answer would be false.