For any two complex numbers $z_1, z_2$,if $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$,then:

The area of the polygon,whose vertices are the non-real roots of the equation $\bar{z} = i z^{2}$ is

Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as:
$S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$
$S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$
$S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$
Then the set $S_{1} \cap S_{2} \cap S_{3}$

If the roots of the equation $Z^3+i Z^2+2 i=0$ are the vertices of a triangle $ABC$,then that triangle $ABC$ is

Let $w = \frac{\sqrt{3} + i}{2}$ and $P = \{w^n : n = 1, 2, 3, \ldots\}$. Further, $H_1 = \{z \in C : \operatorname{Re}(z) > \frac{1}{2}\}$ and $H_2 = \{z \in C : \operatorname{Re}(z) < -\frac{1}{2}\}$, where $C$ is the set of all complex numbers. If $z_1 \in P \cap H_1$, $z_2 \in P \cap H_2$, and $O$ represents the origin, then $\angle z_1 O z_2$ can be:

If ${z_1}$ and ${z_2}$ are complex numbers such that ${z_1} \neq {z_2}$ and $|{z_1}| = |{z_2}|$. If ${z_1}$ has a positive real part and ${z_2}$ has a negative imaginary part,then $\frac{{z_1 + z_2}}{{z_1 - z_2}}$ may be