Let z be a complex number such that the imaginary part of z is non-zero and $$a\, = \,{z^2} + \,z\, + 1$$ is real. Then a cannot take the value

Match the statements in Column I with those in Column II.

[Note : Here z takes value in the complex plane and Im z and Re z denotes, respectively, the imaginary part and the real part of z.]

Column I

(A) The set of points z satisfying $$\left| {z - i} \right|\left. {z\,} \right\|\,\, = \left| {z + i} \right|\left. {\,z} \right\|$$ is contained in or equal to
(B) The set of points z satisfying $$\left| {z + 4} \right| + \,\left| {z - 4} \right| = 10$$ is contained in or equal to
(C) If $$\left| w \right|$$= 2, then the set of points $$z = w - {1 \over w}$$ is contained in or equal to
(D) If $$\left| w \right|$$ = 1, then the set of points $$z = w + {1 \over w}$$ is contained in or equal to.

Column II

(p) an ellipse with eccentricity $${4 \over 5}$$
(q) the set of points z satisfying Im z = 0
(r) the set of points z satisfying $$\left| {{\rm{Im }}\,{\rm{z }}} \right| \le 1$$
(s) the set of points z satisfying $$\,\left| {{\mathop{\rm Re}\nolimits} \,\,z} \right| < 2$$
(t) the set of points z satisfying $$\left| {\,z} \right| \le 3$$

Let $$z = \,\cos \,\theta \, + i\,\sin \,\theta $$ . Then the value of $$\sum\limits_{m = 1}^{15} {{\mathop{\rm Im}\nolimits} } ({z^{2m - 1}})\,at\,\theta \, = {2^ \circ }$$ is