Question
Let $$z$$ and $$\omega $$ be two complex numbers such that $$\left| z \right| \leqslant 1,\left| \omega \right| \leqslant 1\,\,{\text{and }}\left| {z + i\omega } \right| = \left| {z - i\bar \omega } \right| = 2.$$ Then $$z$$ equals
A.
$$1$$ or $$i$$
B.
$$i$$ or $$ - i$$
C.
$$1$$ or $$ - i$$
D.
$$i$$ or $$ - 1$$
Answer :
$$1$$ or $$ - i$$
Solution :
We have, $$2 = \left| {z + i\omega } \right| \leqslant \left| z \right| + \left| \omega \right|\,\,\,\,\,\,\,.....\left( {\text{i}} \right)$$
$$\therefore \,\left| z \right| + \left| \omega \right| \geqslant 2$$
But given that $$\left| z \right| \leqslant 1\,\,{\text{and}}\,\,\left| \omega \right| \leqslant 1\,\,\,\,\,\,\,.....\left( {{\text{ii}}} \right)$$
$$ \Rightarrow \,\left| z \right| + \left| \omega \right| \leqslant 2$$
From (i) and (ii) $$\left| z \right| = \left| \omega \right| = 1$$
$$\eqalign{
& {\text{Also}}\,\,\left| {z + i\omega } \right| = \left| {z - i\bar \omega } \right| \cr
& \Rightarrow \,{\left| {z + i\omega } \right|^2} = {\left| {z - i\bar \omega } \right|^2} \cr
& \Rightarrow \,\left( {z + i\omega } \right)\left( {\bar z - i\bar \omega } \right) = \left( {\bar z + i\omega } \right)\left( {z - i\bar \omega } \right) \cr
& \Rightarrow \,z\bar z + i\omega \bar z - iz\bar \omega + \omega \bar \omega = z\bar z - i\bar z\bar \omega + i\omega z + \omega \bar \omega \cr
& \Rightarrow \,\omega \bar z - \bar \omega z + \bar \omega \bar z - \omega z = 0 \cr
& \Rightarrow \,\left( {\omega + \bar \omega } \right)\left( {\bar z - z} \right) = 0 \cr
& \Rightarrow \,z = \bar z\,\,{\text{or}}\,\,\omega = - \bar \omega \cr
& \Rightarrow \,{I_m}\left( z \right) = 0 \cr
& \Rightarrow \,\operatorname{Re} \left( \omega \right) = 0 \cr
& {\text{Also}}\,\left| z \right| = 1,\left| \omega \right| = 1 \cr
& \Rightarrow \,z = 1\,\,{\text{or}}\,\, - 1\,\,{\text{and}}\,\,\omega = i\,\,{\text{or}}\,\, - i \cr} $$