Explanation

(c)
We have, $\mathcal{z}^{2}\mathcal{+ z}\left| \mathcal{z} \right| + \left| \mathcal{z} \right|^{2} = 0$
$\Rightarrow \ \ \left( \frac{\mathcal{z}}{\left| \mathcal{z} \right|} \right)^{2} + \frac{\mathcal{z}}{\left| \mathcal{z} \right|} + 1 = 0$
This is a quadratic equation in $\frac{\mathcal{z}}{\left| \mathcal{z} \right|}$, therefore roots are $\frac{\mathcal{z}}{\left| \mathcal{z} \right|} = \omega,\omega^{2}\mathcal{\ \Rightarrow \ z =}\omega|\mathcal{z|}$ or $\mathcal{z =}\omega^{2}\mathcal{|z|}$
Let $\mathcal{z =}x + iy$
$\Rightarrow \ \ x + iy = |\mathcal{z|}\left( \frac{- 1}{2} + \frac{i\sqrt{3}}{2} \right)$
or $x + iy = |\mathcal{z|}\left( \frac{- 1}{2} - \frac{i\sqrt{3}}{2} \right)$
$\Rightarrow \ x = - \frac{1}{2}\left| \mathcal{z} \right|,\ y = \left| \mathcal{z} \right|\frac{\sqrt{3}}{2}$
or $x = - \frac{\left| \mathcal{z} \right|}{2},\ y = - \frac{\left| \mathcal{z} \right|\sqrt{3}}{2}$
$\Rightarrow \ y + \sqrt{3}x = 0$
or $y - \sqrt{3}x = 0$
$\Rightarrow \ \ y^{2} - 3x^{2} = 0$
$\Rightarrow$ It represents a pair of straight lines