If the complex number $z=x+iy$,where $i=\sqrt{-1}$,satisfies the condition $|z+1|=1$,then $z$ lies on

The locus of the point $z=x+iy$ satisfying $\left|\frac{z-2i}{z+2i}\right|=1$ is

If $A = \{z = x + iy : \text{real part of } \frac{\bar{z}-1}{z-i} = 2\}$,then the locus of the point $P(x, y)$ in the Cartesian plane is:

Let the complex numbers $z_1, z_2$ and $z_3$ be the vertices of an equilateral triangle. Let $z_0$ be the circumcentre of the triangle,then $z_1^2 + z_2^2 + z_3^2 = $

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 $S = \{z = x + iy : |z - 1 + i| \geq |z|, |z| < 2, |z + i| = |z - 1|\}$. Then the set of all values of $x$,for which $w = 2x + iy \in S$ for some $y \in \mathbb{R}$,is: