$$\eqalign{ & z = \frac{{i{{\left( {\sqrt 3 + i} \right)}^4}}}{{4{{\left( {1 - \sqrt 3 i} \right)}^2}}} = \frac{i}{4} \cdot \frac{{{{\left( {2 + 2\sqrt {3}i } \right)}^2}}}{{ - 2 - 2\sqrt {3}i }} = \frac{{i\left( { - 2 + 2\sqrt {3}i } \right)}}{{2\left( { - 1 - \sqrt {3}i } \right)}} = \frac{{\sqrt 3 + i}}{{1 + \sqrt {3}i }} \cr & z = \frac{{\left( {\sqrt 3 + i} \right)\left( {1 - \sqrt {3}i } \right)}}{{1 + 3}} = \frac{{2\sqrt 3 - 2i}}{4} = \frac{{\sqrt 3 }}{2} - \frac{i}{2} \cr & \therefore \,\,{\text{amp}}\,z = - {\tan ^{ - 1}}\frac{1}{{\sqrt 3 }}. \cr} $$

Rationalizing the given expression
$$\frac{{\left( {2 + 3i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)}}{{1 + 4{{\sin }^2}\theta }}$$
For the given expression to be purely imaginary, real part of the above expression should be equal to zero.
$$\eqalign{ & \Rightarrow \,\,\frac{{2 - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}\theta }} = 0 \cr & \Rightarrow \,\,{\sin ^2}\theta = \frac{1}{3} \cr & \Rightarrow \,\,\sin \theta = \pm \frac{1}{{\sqrt 3 }} \cr} $$

$$\because $$ Real part of roots is 1
Let roots are $$1 + pi , 1 + q$$
∴ sum of roots $$ = 1 + pi + 1 + qi = - \alpha $$     which is real
⇒ $$q = - p$$  or root are
$$1 + pi$$  and $$1 - pi$$  product of roots $$ = 1 + {p^2} = \beta \in \left( {1,\infty } \right)$$
$$p \ne 0$$  as roots are distinct.

$$\eqalign{ & \left| {{z_1} + {z_2}} \right| = \left| {\frac{{{z_1} + {z_2}}}{{{z_1}{z_2}}}} \right| \cr & {\text{or, }}\left| {{z_1} + {z_2}} \right|\left( {1 - \frac{1}{{\left| {{z_1}{z_2}} \right|}}} \right) = 0 \cr & \therefore \,\,\left| {{z_1}{z_2}} \right| = 1. \cr} $$

$$\eqalign{ & z = \frac{{ - 2\left( {1 + 2i} \right)}}{{3 + i}} \cr & = \,\frac{{ - 2 - 4i}}{{3 + i}} = \frac{{ - 2 - 4i}}{{3 + i}} \times \frac{{3 - i}}{{3 - i}} \cr & = \,\frac{{ - 6 + 2i - 12i + 4{i^2}}}{{10}} \cr & = \,\frac{{ - 6 - 10i - 4}}{{10}} = \frac{{ - 10 - 10i}}{{10}} = - 1 - i \cr & z = - 1 - i = r\left( {\cos \,\theta + i\,\sin \,\theta } \right) \cr} $$
On comparing real and imaginary part on both sides, we get
$$\eqalign{ & r\,\cos \,\theta = - 1\,\,\,.....\left( {\text{i}} \right) \cr & r\,\sin \,\theta = - 1\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
On dividing eq. (ii) by (i), we get
$$\eqalign{ & \frac{{r\,\sin \,\theta }}{{r\,\cos \,\theta }} = \frac{{ - 1}}{{ - 1}} \cr & \tan \,\theta = 1 = \tan \,\frac{\pi }{4} \cr & \Rightarrow \,\theta = \frac{\pi }{4} \cr & \therefore \,\theta = \frac{\pi }{4} \cr} $$

$$\eqalign{ & \left| {\frac{{z - 1}}{{z + 1}}} \right| = 1.\,{\text{So, }}\frac{{z - 1}}{{z + 1}} = 1 \cdot \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right) = \frac{{1 + i}}{{\sqrt 2 }}. \cr & \therefore \,\,z = \frac{{\sqrt 2 + 1 + i}}{{\sqrt 2 - 1 - i}} = \frac{{\left( {\sqrt 2 + 1 + i} \right)\left( {\sqrt 2 - 1 + i} \right)}}{{{{\left( {\sqrt 2 - 1} \right)}^2} + 1}},\,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}} \cr} $$

$$\eqalign{ & {z^n} = {\left( {z + 1} \right)^n} \cr & \Rightarrow \,{\left| z \right|^n} = {\left| {z + 1} \right|^n}\,\,{\text{or}}\,\,\,\left| z \right| = \left| {z + 1} \right|. \cr} $$
So the distance of point $$z$$ remain same from $$\left( {0,0} \right)$$  and $$\left( {- 1,0} \right).$$
So, $$z$$ lies on perpendicular bisector of line joining $$\left( {0,0} \right)$$  and $$\left( {-1,0} \right)$$  that is on $$x = - \frac{1}{2}$$

$$\eqalign{ & 2{z^2} + 2z + \lambda = 0 \cr & \Rightarrow \,\,z = \frac{{ - 2 \pm \sqrt {4 - 8\lambda } }}{4}. \cr} $$
For non-real roots $$4 - 8\lambda < 0,{\text{i}}{\text{.e}}{\text{., }}\lambda > \frac{1}{2}.$$

For equilateral triangle,
$$\eqalign{ & \left| {0 - \frac{{ - 1 - i\sqrt {2\lambda - 1} }}{2}} \right| = \left| {0 - \frac{{ - 1 + i\sqrt {2\lambda - 1} }}{2}} \right| = \left| {\frac{{ - 1 + i\sqrt {2\lambda - 1} }}{2} - \frac{{ - 1 - i\sqrt {2\lambda - 1} }}{2}} \right| \cr & {\text{or, }}\frac{1}{4}\left\{ {1 + 2\lambda - 1} \right\} = {\left\{ {\sqrt {2\lambda - 1} } \right\}^2} \cr & \therefore \,\,\frac{\lambda }{2} = 2\lambda - 1 \cr & \therefore \,\,\lambda = \frac{2}{3}. \cr} $$

$$\eqalign{ & \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} + \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} = k\,\,\,\,\,.....\left( 1 \right) \cr & {\text{or, }}{x^2} + {\left( {y - 1} \right)^2} - {x^2} - {\left( {y + 1} \right)^2} = k\left\{ {\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} - \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} } \right\} \cr & \therefore \,\,\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} - \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} = \frac{{ - 4y}}{k}\,\,\,\,\,.....\left( 2 \right) \cr} $$
From (1) and (2),
$$\eqalign{ & 2\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} = k - \frac{{4y}}{k} \cr & \Rightarrow \,\,4{x^2} + \left( {4 - \frac{{16}}{{{k^2}}}} \right){y^2} = {k^2} - 4. \cr} $$
For an ellipse, $$4 - \frac{{16}}{{{k^2}}} > 0,{k^2} - 4 > 0.$$
For $$k = 1, 2$$  the co-efficient of $${y^2}$$ is negative or zero.

$${\left( {b + ia} \right)^5} = {i^5}{\left( {a - ib} \right)^5} = i\left( {\alpha - i\beta } \right).$$