If $i=\sqrt{-1}$, then $\operatorname{Arg}\left[\frac{(1+i)^{2025}}{(1-i)^{2022}}\right]=$

The locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$, where $z=x+i y$, is

If $x_n=\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}$, then $\prod_{n=1}^{\infty} x_n=$

If the roots of the equation $z^2-i=0$ are $\alpha$ and $\beta$, then $|\arg \beta-\arg \alpha|=$