The maximum value of $|z|$ where $z$ satisfies the condition $\left| z + \frac{2}{z} \right| = 2$ is

Let $S_{1}=\{z_{1} \in \mathbb{C}:|z_{1}-3|=\frac{1}{2}\}$ and $S_{2}=\{z_{2} \in \mathbb{C}:|z_{2}-|z_{2}+1||=|z_{2}+|z_{2}-1||\}$. Then,for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$,the least value of $|z_{2}-z_{1}|$ is:

Let $z = x + iy$, where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$ is purely imaginary, lie on

Suppose that $z_{1}, z_{2}, z_{3}$ are three vertices of an equilateral triangle in the Argand plane. Let $\alpha = \frac{1}{2}(\sqrt{3} + i)$ and $\beta$ be a non-zero complex number. The points $\alpha z_{1} + \beta, \alpha z_{2} + \beta, \alpha z_{3} + \beta$ will be

If the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,then the locus of $z=x+iy$ is:

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