Question
If $$z \ne 1\,\,{\text{and }}\frac{{{z^2}}}{{z - 1}}$$ is real, then the point represented by the complex number $$z$$ lies:
A.
either on the real axis or on a circle passing through the origin.
B.
on a circle with center at the origin
C.
either on the real axis or on a circle not passing through the origin.
D.
on the imaginary axis.
Answer :
either on the real axis or on a circle passing through the origin.
Solution :
$$\eqalign{
& {\text{Let }}z = x + iy \cr
& \therefore \,\,{z^2} = {x^2} - {y^2} + 2ixy \cr
& {\text{Now }}\frac{{{z^2}}}{{z - 1}}\,{\text{is real}} \cr
& \Rightarrow \,\,{\text{Im}}\left( {\frac{{{z^2}}}{{z - 1}}} \right) = 0 \cr
& \Rightarrow \,\,{\text{Im}}\left( {\frac{{{x^2} - {y^2} + 2ixy}}{{\left( {x - 1} \right) + iy}}} \right) = 0 \cr
& \Rightarrow \,\,{\text{Im}}\left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left( {x - 1} \right) - iy} \right] = 0 \cr
& \Rightarrow \,\,2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0 \cr
& \Rightarrow \,\,y\left( {{x^2} + {y^2} - 2x} \right) = 0 \cr
& \Rightarrow \,\,y = 0;\,{x^2} + {y^2} - 2x = 0 \cr} $$
∴ $$z$$ lies either on real axis or on a circle through origin.