$${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$

$$ \Rightarrow {\left( i \right)^x} = 1$$

We know $$i = \sqrt { - 1} $$

$$\therefore$$ $${i^2} = - 1$$

$$ \Rightarrow $$ $${i^3} = - 1 \times i = - i$$

$$ \Rightarrow $$ $${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$$

So when power of $$i$$ is 4 or multiple of 4 then it's value is = 1

$$\therefore$$ $${\left( i \right)^x} = 1$$ $$ = {\left( i \right)^{4n}}$$ where n is a positive integer.

Given quadratic equation,
$${Z^2} + aZ + b = 0$$
and two roots are $${Z_1}$$ and $${Z_2}$$.

$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$

Question says,
There are three complex numbers:
1. Origin (0)
2. $${Z_1}$$
3. $${Z_2}$$
and they form an equilateral triangle. So They are the vertices of the triangle.

[ Important Point : If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

In this question,
$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$

By putting those values in the equation we get,

$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${\left( { - a} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${a^2}$$ = $$3b$$

So Option (D) is correct.

[ Note : This question is asked to check if you know the following formula -

"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]

Given that,
$$\left| {z\omega } \right| = 1$$
$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1
$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$       [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$       [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

As $${{z \over \omega }}$$ is imaginary so we can write,

$${{z \over \omega }}$$ = $$ - {{\overline z } \over {\overline \omega }}$$
[ When $$z$$ is imaginary then $$z$$ = $$-\overline z $$ ]

$$ \Rightarrow $$ $$\overline z \omega $$ = $$ - z\overline \omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$$\overline \omega $$.$$\omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$${\left| \omega \right|^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${\left( {{1 \over {\sqrt k }}} \right)^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${1 \over k}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-i$$

Method 2 :

Given that,
$$\left| {z\omega } \right| = 1$$
$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1
$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$       [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$       [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

(1) Magnitude of $$\overline z \omega $$

= $$\left| {\overline z } \right|\left| \omega \right|$$

= $$\left| z \right|\left| \omega \right|$$ [ as $$\left| z \right|$$ = $$\left| {\overline z } \right|$$ ]

= $$\sqrt k $$.$${{1 \over {\sqrt k }}}$$

= 1

$$\therefore$$ The distance from the origin of $${\overline z \omega }$$ is 1.

(2) Argument of $${\overline z \omega }$$ = $$Arg\left( {\overline z \omega } \right)$$

= $$Arg\left( {\overline z } \right) + Arg\left( \omega \right)$$

= $$-Arg\left( z \right) + Arg\left( \omega \right)$$

= $$ - \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)$$

= $$ - {\pi \over 2}$$

$$\therefore$$ $${\overline z \omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\pi \over 2}$$ clockwise.

$$\therefore$$ $${\overline z \omega }$$ = 0 + (-1)$$ \times $$$$i$$ = $$-i$$

Let the circle be $$\left| {z - {z_3}} \right| = r.$$

Then according to given conditions

$$\left| {{z_3} - {z_1}} \right| = r + a$$ (Shown in the image)

and $$\left| {{z_3} - {z_2}} \right| = r + b.$$ (Shown in the image)

Eliminating $$r,$$ we get

$$\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.$$

$$\therefore$$ Locus of center $${z_3}$$ is

$$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b$$ = constant.

Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.

Here distance of z₁ from z₃ is = $$r + a$$ and distance of z₂ from z₃ is = $$r + b$$

Now their difference = ($$r + a$$) - ($$r + b$$) = $$a - b$$ = a constant

$$\therefore$$ Locus of z₃ is a hyperbola.