$$1,\omega $$  and $$\omega^2 $$ are the three cube roots of unity.
$$ \Rightarrow 1 + \omega + {\omega ^2} = 0{\text{ and }}{\omega ^3} = 1.$$
The given expression
$$\eqalign{ & \frac{{a{\omega ^6} + b{\omega ^4} + c{\omega ^2}}}{{b + c{\omega ^{10}} + a{\omega ^8}}} = \frac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}\left[ {{\omega ^6} = 1,{\omega ^4} = \omega } \right] \cr & = \frac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\omega \left( {b + c\omega + a{\omega ^2}} \right)}}\left[ {{\text{Multiplying }}{N^r}{\text{ and }}{D^r}{\text{ by }}\omega {\text{.}}} \right] \cr & = \frac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {a{\omega ^3} + b\omega + c{\omega ^2}} \right)}} = \frac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {a + b\omega + c{\omega ^2}} \right)}} = \omega \cr} $$

$$\eqalign{ & {z_1}{{\bar z}_1} = {z_2}{{\bar z}_2} = ..... = {z_n}{{\bar z}_n} = 1 \cr & \Rightarrow \,{{\bar z}_1} = \frac{1}{{{z_1}}},{{\bar z}_2} = \frac{1}{{{z_2}}},{{\bar z}_3} = \frac{1}{{{z_3}}},.....,{{\bar z}_n} = \frac{1}{{{z_n}}} \cr & \therefore \,\left| {{z_1} + {z_2} + ..... + {z_n}} \right| - \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ..... + \frac{1}{{{z_n}}}} \right| \cr & = \,\left| {{z_1} + {z_2} + ..... + {z_n}} \right| - \left| {{{\bar z}_1} + {{\bar z}_2} + ..... + {{\bar z}_n}} \right| = 0 \cr} $$

$$\left| {{z_1} + {z_2} + {z_3}} \right| = \left| {\left( {{z_1} - 1} \right) + \left( {{z_2} - 2} \right) + \left( {{z_3} - 3} \right) + 6} \right| \leqslant \left| {{z_1} - 1} \right| + \left| {{z_2} - 2} \right| + \left| {{z_3} - 3} \right| + 6 < 1 + 2 + 3 + 6.$$

$$\left| {{z_1}} \right| = $$  the distance of the point representing $${{z_1}}$$ from the origin. So, the distances of the four points from the origin are equal

$$\eqalign{ & \left( {a + ib} \right)\left( {c + id} \right)\left( {e + if} \right)\left( {g + ih} \right) = A + iB\,\,\,\,.....\left( {\text{i}} \right) \cr & \Rightarrow \,\left( {a - ib} \right)\left( {c - id} \right)\left( {e - if} \right)\left( {g - ih} \right) = A - iB\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Multiplying (i) and (ii), we get
$$\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)\left( {{e^2} + {f^2}} \right)\left( {{g^2} + {h^2}} \right) = {A^2} + {B^2}$$

Let $$ABC$$  be the $$\Delta $$ with vertices $$a, b, c$$  and $$PQR$$   be the $$\Delta $$ with vertices $$u, v, w.$$
Then $$c = (1 - r)a + rb$$

$$\eqalign{ & \Rightarrow \,\,c - a = r\left( {b - a} \right) \cr & \Rightarrow \,\,\frac{{c - a}}{{b - a}} = r\,\,\,\,\,......\left( 1 \right) \cr & \Rightarrow \,\,w = \left( {1 - r} \right)u + rv \cr & \Rightarrow \,\,\frac{{w - u}}{{v - u}} = r\,\,\,\,......\left( 2 \right) \cr} $$
From (1) and (2) $$\left| {\frac{{c - a}}{{b - a}}} \right| = \left| {\frac{{w - u}}{{v - u}}} \right|\,\,{\text{and}}$$
$$\eqalign{ & {\text{arg}}\left( {\frac{{c - a}}{{b - a}}} \right) = {\text{arg}}\left( {\frac{{w - u}}{{v - u}}} \right) \cr & \Rightarrow \,\,\frac{{AC}}{{AB}} = \frac{{PR}}{{PQ}}\,\,{\text{and }}\angle CAB = \angle RPQ \cr & \Rightarrow \,\,\Delta ABC \sim \Delta PQR \cr} $$

Here we observe that.
$$AB = AC = AD = 2$$
∴ $$BCD$$  is an arc of a circle with center at $$A$$ and radius 2. Shaded region is outer (exterior) part of this sector $$ABCDA.$$
∴ For any pt. $$z$$ on are $$BCD$$  we should have
$$\left| {z - \left( { - 1} \right)} \right| = 2$$
and for shaded region, $$\left| {z + 1} \right| > 2\,\,\,\,\,\,\,\,......\left( {\text{i}} \right)$$
For shaded region we also have
$$\eqalign{ & - \frac{\pi }{4} < \arg \left( {z + 1} \right) < \frac{\pi }{4} \cr & {\text{or }}\left| {\arg \left( {z + 1} \right)} \right| < \frac{\pi }{4}\,\,\,\,\,\,\,\,\,\,......\left( {{\text{ii}}} \right) \cr} $$
Combining (i) and (ii), (A) is the correct option.

If vertices of a parallelogram are $${z_1},{z_2},{z_3},{z_4}$$   then as diagonals bisect each other
$$\eqalign{ & \therefore \,\,\frac{{{z_1} + {z_3}}}{2} = \frac{{{z_2} + {z_4}}}{2} \cr & \Rightarrow \,\,{z_1} + {z_3} = {z_2} + {z_4} \cr} $$

$$\eqalign{ & x = \frac{{1 \pm \sqrt {3i} }}{2} = \cos \frac{\pi }{3} \pm i\sin \frac{\pi }{3} \cr & \therefore \,\,{x^{2n}} = \cos \frac{{2n\pi }}{3} \pm i\sin \frac{{2n\pi }}{3} \cr & {\text{or,}}\,\,{\left( {{x^n} + \frac{1}{{{x^n}}}} \right)^2} = {x^{2n}} + {x^{ - 2n}} + 2 \cr & {\text{or,}}\,\,{\left( {{x^n} + \frac{1}{{{x^n}}}} \right)^2} = \cos \frac{{2n\pi }}{3} \pm i\sin \frac{{2n\pi }}{3} + \cos \frac{{2n\pi }}{3} \mp i\sin \frac{{2n\pi }}{3} + 2 \cr & {\text{or,}}\,\,{\left( {{x^n} + \frac{1}{{{x^n}}}} \right)^2} = 2 + 2\cos \frac{{2n\pi }}{3}. \cr} $$
∴ the value $$ = 10 + 2\left\{ {\cos \frac{{2\pi }}{3} + \cos \frac{{4\pi }}{3} + \cos \frac{{6\pi }}{3} + \cos \frac{{8\pi }}{3} + \cos \frac{{10\pi }}{3}} \right\}.$$

$$\eqalign{ & \left( {z - i} \right)\left( {{z^2} + 2iz - 2} \right) = 0 \cr & \Rightarrow \,\,z = i,\frac{{ - 2i \pm \sqrt {4{i^2} + 8} }}{2} = i,1 - i, - 1 - i. \cr} $$
Find the area of the triangle whose vertices are $$\left( {0,1} \right),\left( {1, - 1} \right),\left( { - 1, - 1} \right).$$