If z neq 1 and frac{z^2}{z-1} is real,then th | vedclass

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(A) Given that $\frac{z^2}{z-1}$ is real,it must be equal to its conjugate: $\frac{z^2}{z-1} = \overline{\left(\frac{z^2}{z-1}\right)} = \frac{\overli

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either on the real axis or on a circle passing through the origin

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(A) Given that $\frac{z^2}{z-1}$ is real,it must be equal to its conjugate: $\frac{z^2}{z-1} = \overline{\left(\frac{z^2}{z-1}\right)} = \frac{\overline{z}^2}{\overline{z}-1}$.Cross-multiplying,we get: $z^2(\overline{z}-1) = \overline{z}^2(z-1)$.Expanding the terms: $z^2\overline{z} - z^2 = \overline{z}^2z - \overline{z}^2$.Rearranging the terms: $z^2\overline{z} - \overline{z}^2z - z^2 + \overline{z}^2 = 0$.Factoring: $z\overline{z}(z-\overline{z}) - (z^2-\overline{z}^2) = 0$.$z\overline{z}(z-\overline{z}) - (z-\overline{z})(z+\overline{z}) = 0$.$(z-\overline{z})(z\overline{z} - (z+\overline{z})) = 0$.This implies either $z-\overline{z} = 0$ or $z\overline{z} - z - \overline{z} = 0$.$z-\overline{z} = 0$ implies $z$ is purely real (lies on the real axis).$z\overline{z} - z - \overline{z} = 0$ can be written as $(z-1)(\overline{z}-1) = 1$,or $|z-1|^2 = 1$,which represents a circle with center $1$ and radius $1$. This circle passes through the origin since $|0-1| = 1$.Thus,$z$ lies on the real axis or on a circle passing through the origin.

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

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