$$z = {\left( {\cos \theta + i\sin \theta } \right)^2} = \cos 2\theta + i\sin 2\theta ,\,{\text{where }}\frac{\pi }{2} < 2\theta < \pi .$$
∴ $$z$$ is a complex number in the second quadrant. So, $$\frac{\pi }{2} < \arg \,z < \pi .$$
$$\therefore \,\,\arg\,z = {\tan ^{ - 1}}\left( {\tan 2\theta } \right) = 2\theta .$$

Expression $$ = \frac{{\omega \left( {2 + 3\omega + {\omega ^2}} \right)}}{{2 + 3\omega + {\omega ^2}}} + \frac{{\omega \left( {3 + \omega + 2{\omega ^2}} \right)}}{{3 + \omega + 2{\omega ^2}}} = \omega + \omega .$$

Given complex number is
$$\eqalign{ & \left( {1 - \sin \theta } \right) + i\,\cos \theta \equiv a + ib \cr & {\text{Argument}} \equiv {\text{tan}}\,\theta = \frac{b}{a} \cr & \Rightarrow \,\tan \theta = \frac{{\cos \theta }}{{1 - \sin \theta }} \cr & = \,\frac{{{{\cos }^2}\frac{\theta }{2} - {{\sin }^2}\frac{\theta }{2}}}{{{{\sin }^2}\frac{\theta }{2} + {{\cos }^2}\frac{\theta }{2} - 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}} \cr & = \,\frac{{\left( {\cos \frac{\theta }{2} - \sin \frac{\theta }{2}} \right)\left( {\cos \frac{\theta }{2} + \sin \frac{\theta }{2}} \right)}}{{{{\left( {\sin \frac{\theta }{2} - \cos \frac{\theta }{2}} \right)}^2}}} \cr & = \,\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} - \sin \frac{\theta }{2}}} \cr & = \,\frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}} = \frac{{\tan \frac{\pi }{4} + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\pi }{4}\tan \frac{\theta }{2}}} \cr & \tan \,\theta = \tan \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) \cr & {\text{Hence, argument}} = \frac{\pi }{4} + \frac{\theta }{2} \cr} $$

$$\left| z \right| + \left| {z - 2} \right| \geqslant 2$$

$$\eqalign{ & {\left| {1 - {{\bar z}_1}{z_2}} \right|^2} - {\left| {{z_1} - {z_2}} \right|^2} \cr & = \left( {1 - {{\bar z}_1}{z_2}} \right)\left( {1 - {z_1}{{\bar z}_2}} \right) - \left( {{z_1} - {z_2}} \right)\left( {{{\bar z}_1} - {{\bar z}_2}} \right) \cr & = 1 - {z_1}{{\bar z}_2} - {{\bar z}_1}{z_2} + {z_1}{{\bar z}_1}{z_2}{{\bar z}_2} - \left( {{z_1}{{\bar z}_1} - {z_1}{{\bar z}_2} - {{\bar z}_1}{z_2} + {z_2}{{\bar z}_2}} \right) \cr & = 1 + {z_1}{{\bar z}_1}{z_2}{{\bar z}_2} - {z_1}{{\bar z}_1} - {z_2}{{\bar z}_2} \cr & = 1 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} - {\left| {{z_1}} \right|^2} - {\left| {{z_2}} \right|^2} \cr & = \left( {1 - {{\left| {{z_1}} \right|}^2}} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) \cr & \Rightarrow k = 1 \cr} $$

$$\eqalign{ & {\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n} \cr & \Rightarrow \,\,{\left( { - \omega } \right)^n} = {\left( {1 + \omega } \right)^n} = {\left( { - {\omega ^2}} \right)^n} \cr & \Rightarrow \,\,{\omega ^n} = 1 \cr & \Rightarrow \,\,n = 3 \cr} $$

Expression $$ = \left( {1 + i} \right)2i + {\left( { - 2i} \right)^3}.$$

$${\text{Since}}\,\,a{z^2} + bz + c = 0\,\,\,.....\left( 1 \right)$$
and $$z_1 , z_2$$  (roots of (1)) are such that $$\operatorname{Im} \left( {{z_1},{z_2}} \right) \ne 0.$$
Now, $$z_1$$ and $$z_2$$ are not conjugates of each other
Complex roots of (1) are not conjugate of each other
Co-efficient $$a, b, c$$  can-not all be real at least one $$a, b, c$$  is imaginary.

$$\eqalign{ & {\text{ATQ }}\left| {x + iy - 5i} \right| = \left| {x + iy + 5i} \right| \cr & \Rightarrow \,\,\left| {x + \left( {y - 5} \right)i} \right| = \left| {x + \left( {y + 5} \right)i} \right| \cr & \Rightarrow \,\,{x^2} + {\left( {y - 5} \right)^2} = {x^2} + {\left( {y + 5} \right)^2} \cr & \Rightarrow \,\,{x^2} + {y^2} - 10y + 25 = {x^2} + {y^2} + 10y + 25 \cr & \Rightarrow \,\,20y = 0 \cr & \Rightarrow \,\,y = 0 \cr} $$
$$\therefore \,\,'A'$$   is the correct alternative.

We have, $$2 = \left| {z + i\omega } \right| \leqslant \left| z \right| + \left| \omega \right|\,\,\,\,\,\,\,.....\left( {\text{i}} \right)$$
$$\therefore \,\left| z \right| + \left| \omega \right| \geqslant 2$$
But given that $$\left| z \right| \leqslant 1\,\,{\text{and}}\,\,\left| \omega \right| \leqslant 1\,\,\,\,\,\,\,.....\left( {{\text{ii}}} \right)$$
$$ \Rightarrow \,\left| z \right| + \left| \omega \right| \leqslant 2$$
From (i) and (ii) $$\left| z \right| = \left| \omega \right| = 1$$
$$\eqalign{ & {\text{Also}}\,\,\left| {z + i\omega } \right| = \left| {z - i\bar \omega } \right| \cr & \Rightarrow \,{\left| {z + i\omega } \right|^2} = {\left| {z - i\bar \omega } \right|^2} \cr & \Rightarrow \,\left( {z + i\omega } \right)\left( {\bar z - i\bar \omega } \right) = \left( {\bar z + i\omega } \right)\left( {z - i\bar \omega } \right) \cr & \Rightarrow \,z\bar z + i\omega \bar z - iz\bar \omega + \omega \bar \omega = z\bar z - i\bar z\bar \omega + i\omega z + \omega \bar \omega \cr & \Rightarrow \,\omega \bar z - \bar \omega z + \bar \omega \bar z - \omega z = 0 \cr & \Rightarrow \,\left( {\omega + \bar \omega } \right)\left( {\bar z - z} \right) = 0 \cr & \Rightarrow \,z = \bar z\,\,{\text{or}}\,\,\omega = - \bar \omega \cr & \Rightarrow \,{I_m}\left( z \right) = 0 \cr & \Rightarrow \,\operatorname{Re} \left( \omega \right) = 0 \cr & {\text{Also}}\,\left| z \right| = 1,\left| \omega \right| = 1 \cr & \Rightarrow \,z = 1\,\,{\text{or}}\,\, - 1\,\,{\text{and}}\,\,\omega = i\,\,{\text{or}}\,\, - i \cr} $$