JEE MAIN - Mathematics (2012 - No. 21)
If $$z \ne 1$$ and $$\,{{{z^2}} \over {z - 1}}\,$$ is real, then the point represented by the complex number z lies :
either on the real axis or a circle passing through the origin.
on a circle with centre at the origin
either on real axis or on a circle not passing through the origin.
on the imaginary axis.
Explanation
Let $$z = x + iy$$
$$\therefore$$ $$\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$$
Now $${{{z^2}} \over {z - 1}}$$ is real
$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$$
$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$$
$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$$
$$ \Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0$$
$$ \Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0$$
$$ \Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$$
$$\therefore$$ $$\,\,\,\,$$ $$z$$ lies either on real axis or on a circle through origin.
$$\therefore$$ $$\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$$
Now $${{{z^2}} \over {z - 1}}$$ is real
$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$$
$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$$
$$ \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$$
$$ \Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0$$
$$ \Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0$$
$$ \Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$$
$$\therefore$$ $$\,\,\,\,$$ $$z$$ lies either on real axis or on a circle through origin.
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