If $$z_1=2+3 i$$ and $$z_2=3+2 i$$, where $$i=\sqrt{-1}$$, then $$\left[\begin{array}{cc}z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1\end{array}\right]\left[\begin{array}{cc}\bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1\end{array}\right]$$ is equal to

The radius of the circle represented by $$(1+i)(1+3i)(1+7i)=x+iy$$ is $$(i=\sqrt{-1})$$.

If $$1, \alpha_1, \alpha_2, \alpha_3$$ and $$\alpha_4$$ are the roots of $$z^5-1=0$$ and $$\omega$$ is a cube root of units, then $$(\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-\alpha_3\right)\left(\omega-\alpha_4\right)+\omega$$ is equal to

If $$a > 0$$ and $$z=x+i y$$, then $$\log _{\cos ^2 \theta}|z-a|>\log _{\cos ^2 \theta}|z-a i|,(\theta \in R)$$ implies