A real value of $$x$$ will satisfy the equation, $$\left(\frac{3-4 i x}{3+4 i x}\right)=\alpha-i \beta,(\alpha, \beta$$ are real $$)$$, if

What is the value of $$(1-i \sqrt{3})^9$$ is equal to

$$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020}$$ is equal to

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