Question
If $$\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1,$$ then $$z$$ lies on
A.
an ellipse
B.
the imaginary axis
C.
a circle
D.
the real axis
Answer :
the imaginary axis
Solution :
$$\eqalign{
& \left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1 \cr
& \Rightarrow \,\,{\left| {{z^2} - 1} \right|^2} = {\left( {z\overline z + 1} \right)^2} \cr
& \Rightarrow \,\,\left( {{z^2} - 1} \right)\left( {{{\overline z }^2} - 1} \right) = {\left( {z\overline z + 1} \right)^2} \cr
& \Rightarrow \,\,{z^2}{\overline z ^2} - {z^2} - {\overline z ^2} + 1 = \,{z^2}{\overline z ^2} + 2z\overline z + 1 \cr
& \Rightarrow \,\,{z^2} + 2z\overline z + {\overline z ^2} = 0 \cr
& \Rightarrow \,\,{\left( {z + \overline z } \right)^2} = 0 \cr
& \Rightarrow \,\,z = - \overline z \cr
& \Rightarrow \,\,z\,\,{\text{is purely imaginary}} \cr} $$