Let the complex number $$z = x + iy$$ be such that $${{2z - 3i} \over {2z + i}}$$ is purely imaginary. If $${x} + {y^2} = 0$$, then $${y^4} + {y^2} - y$$ is equal to :

Let $$A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.$$ is purely imaginary $$\}$$. Then the sum of the elements in $$\mathrm{A}$$ is :

If for $$z=\alpha+i \beta,|z+2|=z+4(1+i)$$, then $$\alpha+\beta$$ and $$\alpha \beta$$ are the roots of the equation :

Let $$a \neq b$$ be two non-zero real numbers. Then the number of elements in the set $$X=\left\{z \in \mathbb{C}: \operatorname{Re}\left(a z^{2}+b z\right)=a\right.$$ and $$\left.\operatorname{Re}\left(b z^{2}+a z\right)=b\right\}$$ is equal to :