Question
If $$z = x + iy\,\,{\text{and }}\omega = \frac{{\left( {1 - iz} \right)}}{{\left( {z - i} \right)}},{\text{then }}\left| \omega \right| = 1$$ implies that, in the complex plane,
A.
$$z$$ lies on the imaginary axis
B.
$$z$$ lies on the real axis
C.
$$z$$ lies on the unit circle
D.
None of these
Answer :
$$z$$ lies on the real axis
Solution :
$$\eqalign{
& \left| \omega \right| = 1 \cr
& \Rightarrow \,\,\left| {\frac{{1 - iz}}{{z - i}}} \right| = 1 \cr
& \Rightarrow \,\,\left| {1 - iz} \right| = \left| {z - i} \right| \cr
& \Rightarrow \,\,\left| {1 - i\left( {x + iy} \right)} \right| = \left| {x + iy - i} \right| \cr
& \Rightarrow \,\,\left| {\left( {y + 1} \right) - ix} \right|\, = \left| {x + i\left( {y - 1} \right)} \right| \cr
& \Rightarrow \,\,{x^2} + {\left( {y + 1} \right)^2} = {x^2} + {\left( {y - 1} \right)^2} \cr
& \Rightarrow \,\,4y = 0 \cr
& \Rightarrow \,\,y = 0 \cr
& \Rightarrow \,\,z\,\,{\text{lies on real axis}}\, \cr} $$