Given that $$\left| z \right| = 1\,\,{\text{and }}\omega = \frac{{z - 1}}{{z + 1}}\left( {z \ne - 1} \right)$$
Now we know that $$z\overline z = {\left| z \right|^2}$$
$$\eqalign{ & \Rightarrow \,\,z\overline z = 1\,\,\,\,\,\,\,\left( {{\text{for }}\left| z \right| = 1} \right) \cr & \therefore \,\,\omega = \left( {\frac{{z - 1}}{{z + 1}}} \right) \times \frac{{\left( {\overline z + 1} \right)}}{{\left( {\overline z + 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{z\overline z + z - \overline z - 1}}{{z\overline z + z + \overline z + 1}} = \frac{{2iy}}{{2 + 2x}} \cr & \left[ {\because \,\,z\overline z = 1\,\,{\text{and taking }}z = x + i\,y\,\,\,{\text{so that}}\,\,z + \overline z = 2x\,\,{\text{and }}z - \overline z = 2i\,y} \right] \cr & \Rightarrow \,\,{\text{Re}}\left( \omega \right) = 0 \cr} $$

$$\eqalign{ & {\left[ {\frac{{\sin \frac{\pi }{6} + i\left( {1 - \cos \frac{\pi }{6}} \right)}}{{\sin \frac{\pi }{6} - i\left( {1 - \cos \frac{\pi }{6}} \right)}}} \right]^3} \cr & = {\left[ {\frac{{2\sin \frac{\pi }{{12}}\cos \frac{\pi }{{12}} + i\left( {2{{\sin }^2}\frac{\pi }{{12}}} \right)}}{{2\sin \frac{\pi }{{12}}\cos \frac{\pi }{{12}} - i\left( {2{{\sin }^2}\frac{\pi }{{12}}} \right)}}} \right]^3} \cr & = {\left[ {\frac{{\cos \frac{\pi }{{12}} + i\sin \frac{\pi }{{12}}}}{{\cos \frac{\pi }{{12}} - i\sin \frac{\pi }{{12}}}}} \right]^3} \cr & = {\left( {\frac{{{e^{i\frac{\pi }{{12}}}}}}{{{e^{ - i\frac{\pi }{12}}}}}} \right)^3} \cr & = {\left( {{e^{i\frac{\pi }{6}}}} \right)^3} \cr & = {e^{i \times 3 \times \frac{\pi }{6}}} \cr & = {e^{i\frac{\pi }{2}}} \cr & = \cos \frac{\pi }{2} + i\sin \frac{\pi }{2} \cr & = i \cr} $$

$$\eqalign{ & \sin \left\{ {\left( {{\omega ^{13}} + {\omega ^2}} \right)\pi + \frac{\pi }{4}} \right\} \cr & = \sin \left\{ {\left( {\omega + {\omega ^2}} \right)\pi + \frac{\pi }{4}} \right\} = \sin \left( { - \pi + \frac{\pi }{4}} \right) \cr & = - \sin \frac{\pi }{4} = - \frac{1}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & {\tan ^{ - 1}}\frac{y}{{x - 1}} = {\tan ^{ - 1}}\frac{{y + 3}}{x} \cr & \therefore \,\,xy = \left( {x - 1} \right)\left( {y + 3} \right) \cr & \Rightarrow \,\,3\left( {x - 1} \right) = y. \cr} $$

$$\eqalign{ & {z^2} + z + 1 = 0 \cr & \Rightarrow z = \omega \,\,{\text{or }}{\omega ^2} \cr & {\text{So, }}z + \frac{1}{z} = \omega + {\omega ^2} = - 1 \cr & {z^2} + \frac{1}{{{z^2}}} = {\omega ^2} + \omega = - 1, \cr & {z^3} + \frac{1}{{{z^3}}} = {\omega ^3} + {\omega ^3} = 2 \cr & {z^4} + \frac{1}{{{z^4}}} = - 1,{z^5} + \frac{1}{{{z^5}}} = - 1{\text{ and }}{z^6} + \frac{1}{{{z^6}}} = 2 \cr} $$
∴ The given sum $$ = 1 + 1 + 4 +1 +1 + 4 = 12$$

If in a complex number $$a + ib,$$   the ratio $$a : b$$  is $$1:\sqrt 3 ,$$  then it always convert the complex number in $$\omega .$$
$$\eqalign{ & {\text{Since, }}\omega = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i \cr & \therefore \,4 + 5{\left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right)^{365}} \cr & = 4 + 5{\omega ^{334}} + 3{\omega ^{365}} \cr & = 4 + 5 \cdot {\left( {{\omega ^3}} \right)^{111}} \cdot \omega + 3 \cdot {\left( {{\omega ^3}} \right)^{121}} \cdot {\omega ^2} \cr & = 4 + 5\omega + 3{\omega ^2}\,\,\,\,\,\,\,\,\,\left( {\because \,{\omega ^3} = 1} \right) \cr & = 1 + 3 + 2\omega + 3\omega + 3{\omega ^2} \cr & = 1 + 2\omega + 3\left( {1 + \omega + {\omega ^2}} \right) \cr & = 1 + 2\omega + 3 \times 0\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \,1 + \omega + {\omega ^2} = 0} \right) \cr & = 1 + \left( { - 1 + \sqrt 3 i} \right) \cr & = \sqrt 3 i \cr} $$

$$\eqalign{ & \frac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}} = \frac{{1 - i\sqrt 3 }}{2} \cr & \Rightarrow \,\,\arg \left( {\frac{{{z_1} - {z_3}}}{{{z_{_2}} - {z_3}}}} \right) = \arg \left( {\frac{{1 - i\sqrt 3 }}{2}} \right) \cr & \Rightarrow \,\,\arg \left( {\cos \left( { - \frac{\pi }{3}} \right) + i\sin \left( { - \frac{\pi }{3}} \right)} \right) \cr & \Rightarrow \,\,{\text{angle}}\,{\text{between}}\,{\text{ }}{z_1} - {z_3}\,{\text{and}}\,{z_2} - {z_3}{\text{ is }}{60^ \circ }. \cr & {\text{and }}\left| {\frac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}}} \right| = \left| {\frac{{1 - i\sqrt 3 }}{2}} \right| \cr & \Rightarrow \,\,\left| {\frac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}}} \right| = 1 \cr} $$
$$ \Rightarrow \,\,\left| {{z_1} - {z_3}} \right| = \left| {{z_2} - {z_3}} \right|$$         NOTE THIS STEP
⇒ The $$\Delta $$ with vertices $${z_1},{z_2}{\text{ and }}{z_3}$$   is isosceles with vertical $$\angle {60^ \circ }.$$  Hence rest of the two angles should also be 60° each.
⇒ Req. $$\Delta $$ is an equilateral $$\Delta .$$

$$\eqalign{ & {x^2} + {y^2} + \left( {4 - 3i} \right)\left( {x + iy} \right) + \left( {4 + 3i} \right)\left( {x - iy} \right) + 5 = 0 \cr & {\text{or, }}{x^2} + {y^2} + 8x + 6y + 5 = 0 \cr} $$
∴ radius $$ = \sqrt {{4^2} + {3^2} - 5} = \sqrt {20} = 2\sqrt 5 .$$

$$z$$ lies on or inside the circle with center $$( - 4, 0)$$  and radius 3 units.

From the Argand diagram maximum value of $$\left| {z + 1} \right|$$   is 6

As given $$w = \frac{z}{{z - \frac{1}{3}i}}$$
$$ \Rightarrow \,\,\left| w \right| = \frac{{\left| z \right|}}{{\left| {z - \frac{1}{3}i} \right|}} = 1$$
⇒ distance of $$z$$ from origin and point $$\left( {0,\frac{1}{3}} \right)$$  is same hence $$z$$ lies on bisector of the line joining points (0, 0) and $$\left( {0,\frac{1}{3}} \right).$$
Hence $$z$$ lies on a straight line.