If z1 and z^2z-1 is real, then the point represented by the complex number z lies:

Question

If $z
e1$ and $\dfrac{z^2}{z-1}$ is real, then the point represented by the complex number $z$ lies:

Accepted Answer

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.

Solutions · 1