Given $$\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta $$

As we know, $$\,\,\,\,\overrightarrow z = {1 \over z}$$

$$\therefore$$ $$\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$$

$$ = \arg \left( z \right) = \theta .$$

Let $$z = x + iy$$

$$\therefore$$ $$\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$$

Now $${{{z^2}} \over {z - 1}}$$ is real

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$$

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$$

$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$$

$$ \Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0$$

$$ \Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0$$

$$ \Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$$

$$\therefore$$ $$\,\,\,\,$$ $$z$$ lies either on real axis or on a circle through origin.

$${\left( {1 + \omega } \right)^7} = A + B\omega ;\,\,\,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega $$

$$ - {\omega ^2} = A + B\omega ;\,\,\,\,\,\,\,\,\,\,1 + \omega = A + B\omega $$

$$ \Rightarrow A = 1,B = 1.$$

As real part of roots is $$1$$

Let roots are $$1 + pi,1 + q$$

$$\therefore$$ sum of roots $$ = 1 + pi + 1 + qi = - \alpha $$

which is real $$ \Rightarrow q = - p\,\,$$

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.