A) either on the real axis or on a circle passing through the origin.
B) on a circle with centre at the origin.
C) either on the real axis or on a circle not passing through the origin.
D) on the imaginary axis.
Correct Answer: A
Solution :
\[\frac{{{z}^{2}}}{z-1}=\frac{{{\overline{z}}^{2}}}{z-1}\] \[\Rightarrow z\overline{z}z-{{z}^{2}}=z\overline{z}\overline{z}-{{z}^{2}}\] \[\Rightarrow {{\left| z \right|}^{2}}(z-\overline{z})-(z-\overline{z})(z+\overline{z})=0\] \[\Rightarrow (z-\overline{z})\,\,({{\left| z \right|}^{2}}-(z+\overline{z})=0\] Either \[z=\overline{z}\Rightarrow \] real axis or \[{{\left| z \right|}^{2}}=z+z\Rightarrow z\overline{z}-z-\overline{z}=0\] represents a circle passing through origin.
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