$$\left| {{{{z_1} - 2{z_2}} \over {2 - {z_1}{{\overline z }_2}}}} \right| = 1 \Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2}$$

$$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right)$$

$$ \Rightarrow \left( {{z_1} - 2{z_1}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - {{\overline z }_1}{z_2}} \right)$$

$$ \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2}$$

$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2}$$

$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0$$

$$\left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0$$

As $$\,\,\,\left| {{z_2}} \right| \ne 1\,\,\,$$ $$\therefore$$ $$\,\,\,{\left| {{z_1}} \right|^2} = 4 \Rightarrow \left| {{z_1}} \right| = 2$$

$$ \Rightarrow \,$$ Point $$\,{z_1}\,$$ lies on circle of radius $$2.$$

We know minimum value of

$$\,\,\,\left| {{Z_1} + {Z_2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|$$

Thus minimum value of

$$\,\,\,\left| {Z + {1 \over 2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| Z \right| - {1 \over 2}} \right|$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \left| {Z + {1 \over 2}} \right| \le \left| Z \right| + {1 \over 2}$$

Since, $$\,\,\,\left| Z \right| \ge 2$$

$$\therefore$$ $$\,\,\,2 - {1 \over 2} < \left| {Z + {1 \over 2}} \right| < 2 + {1 \over 2}$$

$$ \Rightarrow {3 \over 2} < \left| {Z + {1 \over 2}} \right| < {5 \over 2}$$

Given $$\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta $$

As we know, $$\,\,\,\,\overrightarrow z = {1 \over z}$$

$$\therefore$$ $$\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$$

$$ = \arg \left( z \right) = \theta .$$

Let $$z = x + iy$$

$$\therefore$$ $$\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$$

Now $${{{z^2}} \over {z - 1}}$$ is real

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$$

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$$

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$$

$$ \Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0$$

$$ \Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0$$

$$ \Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$$

$$\therefore$$ $$\,\,\,\,$$ $$z$$ lies either on real axis or on a circle through origin.