The locus of $z$ satisfying $|z|+|z-1|=3$ is

Let $C$ be the set of all complex numbers. Let $S_{1}=\{z \in C:|z-2| \leq 1\}$ and $S_{2}=\{z \in C: z(1+i)+\overline{z}(1-i) \geq 4\}$. Then,the maximum value of $\left|z-\frac{5}{2}\right|^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to:

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:

The triangle formed by the complex numbers $z_1$,$z_2$,and $-\omega z_1 - \omega^2 z_2$ on the Argand plane is:

If $z_1, z_2, z_3$ are points in the Argand plane,then $\left| \begin{array}{ccc} z_1 & \overline{z_1} & 1 \\ z_2 & \overline{z_2} & 1 \\ z_3 & \overline{z_3} & 1 \end{array} \right| = $

Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}$,where $z \in \mathbb{C}$,be the equation of a circle with center at $C$. If the area of the triangle,whose vertices are at the points $(0,0)$,$C$,and $(\alpha, 0)$,is $11$ square units,then $\alpha^2$ equals