Let $$S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}$$ and $$S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}$$. Then, for $$z_{1} \in S_{1}$$ and $$z_{2} \in S_{2}$$, the least value of $$\left|z_{2}-z_{1}\right|$$ is :

Let S be the set of all $$(\alpha, \beta), \pi<\alpha, \beta<2 \pi$$, for which the complex number $$\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$$ is purely imaginary and $$\frac{1+i \cos \beta}{1-2 i \cos \beta}$$ is purely real. Let $$Z_{\alpha \beta}=\sin 2 \alpha+i \cos 2 \beta,(\alpha, \beta) \in S$$. Then $$\sum\limits_{(\alpha, \beta) \in S}\left(i Z_{\alpha \beta}+\frac{1}{i \bar{Z}_{\alpha \beta}}\right)$$ is equal to :

Let the minimum value $$v_{0}$$ of $$v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}$$ is attained at $${ }{z}=z_{0}$$. Then $$\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$$ is equal to

If $$z=x+i y$$ satisfies $$|z|-2=0$$ and $$|z-i|-|z+5 i|=0$$, then