Let $z$ be a purely imaginary number such that $\text{Im}(z) < 0$. Then $\arg(z)$ is equal to

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\overline{z}_{1} = i \overline{z}_{2}$ and $\arg \left( \frac{z_{1}}{\overline{z}_{2}} \right) = \pi$. Then:

$\operatorname{Arg}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)$ lies in the interval

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|,$ then $\text{arg}({z_1}) - \text{arg}({z_2})$ is equal to

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

If $z=x+iy$,where $x, y \in \mathbb{R}$ and the point $P$ in the Argand plane represents $z$,then the locus of $P$ satisfying the condition $\arg \left(\frac{z-1}{z-3i}\right)=\frac{\pi}{2}$ is: