Let $$z$$ and $$\omega $$ be two complex numbers such that
$$\left| z \right| \le 1,$$ $$\left| \omega \right| \le 1$$ and $$\left| {z + i\omega } \right| = \left| {z - i\overline \omega } \right| = 2$$ then $$z$$ equals

$${\rm{z }} \ne {\rm{0}}$$ is a complex number

Column I

(A) Re z = 0
(B) Arg $$z = {\pi \over 4}$$

Column II

(p) Re$${z^2}$$ = 0
(q) Im$${z^2}$$ = 0
(r) Re$${z^2}$$ = Im$${z^2}$$

If $$a,\,b,\,c$$ and $$u,\,v,\,w$$ are complex numbers representing the vertics of two triangles such that $$c = \left( {1 - r} \right)a + rb$$ and $$w = \left( {1 - r} \right)u + rv,$$ where $$w = \left( {1 - r} \right)u + rv,$$ is a complex number, then the two triangles

If $$z = x + iy$$ and $$\omega = \left( {1 - iz} \right)/\left( {z - i} \right),$$ then $$\,\left| \omega \right| = 1$$ implies that, in the complex plane,