Question
Let $$z$$ and $$\omega $$ be two complex numbers such that $$\left| z \right| \leqslant 1,\left| \omega \right| \leqslant 1\,\,{\text{and }}\left| {z + i\omega } \right| = \left| {z - i\overline \omega } \right| = 2$$ then $$z$$ equals
A.
1 or $$i$$
B.
$$i$$ or $$- i$$
C.
1 or $$- 1$$
D.
$$i$$ or $$- 1$$
Answer :
1 or $$- 1$$
Solution :
$$\eqalign{
& {\text{Given that }}\left| {z + i\omega } \right| = \left| {z - i\overline \omega } \right| \cr
& \Rightarrow \,\,\left| {z - \left( { - i\omega } \right)} \right| = \left| {z - \left( { - \overline {i\omega } } \right)} \right| \cr} $$
⇒ $$z$$ lies on perpendicular bisector of the line segment joining $${\left( { - i\omega } \right)}$$ and $${\left( { - \overline {i\omega } } \right)}$$ , which is real axis,
$${\left( { - i\omega } \right)}\,$$ and $${\left( { - \overline {i\omega } } \right)}\,$$ being mirror images of each other.
$$\eqalign{
& \therefore \,\,{\text{Im}}\left( z \right) = 0. \cr
& {\text{If }}z = x\,\,{\text{then }}\left| z \right| \leqslant 1 \cr
& \Rightarrow \,\,{x^2} \leqslant 1 \cr
& \Rightarrow \,\, - 1 \leqslant x \leqslant 1 \cr} $$
∴ $$(C)$$ is the correct option.