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Let z$_1$ and z$_2$ be two complex numbers such that $\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}$ is unimodular. If z$_2$ is not unimodular then $|{z_1}| = 2$.

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Let us first get to know about what is $z$ and $\bar z$.

$z$ is a complex number which has a real part and an imaginary part. For example:- say $z = x + iy$, then $x$ is the real part and $y$ is the imaginary part.

$\bar z$ is known to be conjugate of $z$, which means that its real part remains as same as of the $z$, but the imaginary part becomes negative. So, $\bar z = x - iy$.

Let us now use the fact that $z.\bar z = |z{|^2}$……(1)

We are already given that z$_2$ is not unimodular. So, |z$_2$|$^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 $z$ is unimodular, then $|z| = 1$.

Since, we are given that $\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}$ is unimodular.

So, $\dfrac{{|{z_1} - 2{z_2}|}}{{|2 - {z_1}{{\bar z}_2}|}} = 1$.

$\therefore |{z_1} - 2{z_2}| = |2 - {z_1}{\bar z_2}|$

Squaring both sides, we will get:-

$|{z_1} - 2{z_2}{|^2} = |2 - {z_1}{\bar z_2}{|^2}$

Now using (1), we have:-

$({z_1} - 2{z_2})({\bar z_1} - 2{\bar z_2}) = (2 - {z_1}{\bar z_2})(2 - {\bar z_1}{z_2})$

Simplifying both the sides, we will get:-

${z_1}{\bar z_1} - 2{\bar z_1}{z_2} - 2{z_1}{\bar z_2} + 4{z_2}{\bar z_2} = 4 - 2{z_1}{\bar z_2} - 2{\bar z_1}{z_2} + {z_1}{\bar z_1}{z_2}{\bar z_2}$

Now using (1) and cancelling out the common terms from both the sides:-

$|{z_1}{|^2} + 4|{z_2}{|^2} = 4 - |{z_1}{|^2}|{z_2}{|^2}$

$|{z_1}{|^2} + 4|{z_2}{|^2} + |{z_1}{|^2}|{z_2}{|^2} - 4 = 0$

\[(|{z_2}{|^2} - 1) + (|{z_1}{|^2} - 4) = 0\]

So, either \[(|{z_2}{|^2} - 1) = 0\] or \[(|{z_1}{|^2} - 4) = 0\]

Using (2), we see that the only possibility is:

\[(|{z_1}{|^2} - 4) = 0\] …….(3)

Let ${z_1} = x + iy$

Since, \[|{z_1}| = \sqrt {{x^2} + {y^2}} \]

Hence, \[|{z_1}{|^2} = {\left( {\sqrt {{x^2} + {y^2}} } \right)^2} = {x^2} + {y^2}\]

Putting this in (3), we will get:-

(\[{x^2} + {y^2}\] - 4) = 0

So, \[{x^2} + {y^2}\] = 4.

This equation depicts a circle with center (0, 0) and radius 2.

Because the general equation of a circle is given by ${(x - h)^2} + {(y - k)^2} = {r^2}$, where $(h,k)$ is the center of the circle and $r$ is the radius.

Hence, ${z_1}$ lies on a circle and ${z_1}$ = 2.

Hence, the statement given is true.

Also remember the fact that $z.\bar z = |z{|^2}$.

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 z$_1$ and z$_2$ be two complex numbers such that $\dfrac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}$ is unimodular. If z$_2$ is not unimodular then $|{z_1}| \ne $ 2.

In this case our answer would be false.