$\frac{{\mathrm{z}}^2}{\mathrm{z}-1}\mathrm{is}\mathrm{real}$

Let z = x + iy

 $\Rightarrow \frac{{\mathrm{z}}^2}{\mathrm{z}-1}=\frac{{\mathrm{x}}^2-{\mathrm{y}}^2+2\text{i}\mathrm{xy}}{(\mathrm{x}-1)+\text{i}\mathrm{y}}=\frac{({\mathrm{x}}^2-{\mathrm{y}}^2+2\text{i}\mathrm{xy})(\mathrm{x}-1-\text{i}\mathrm{y})}{{(\mathrm{x}-1)}^2+{\mathrm{y}}^2}$

Now, imaginary part of  $\frac{z^2}{z-1}$ equal to zero.

$\Rightarrow$ – y (x² – y²) + (x – 1) 2xy = 0

$\Rightarrow$ y (x² + y² – 2x) = 0

$\Rightarrow$ y = 0 or x² + y² – 2x = 0.