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 :

Let $$a,b$$ be two real numbers such that $$ab < 0$$. IF the complex number $$\frac{1+ai}{b+i}$$ is of unit modulus and $$a+ib$$ lies on the circle $$|z-1|=|2z|$$, then a possible value of $$\frac{1+[a]}{4b}$$, where $$[t]$$ is greatest integer function, is :