If a complex number $z$ satisfies $|z^2-1|=|z|^2+1$,then $z$ lies on

Let $C$ be the circle in the complex plane with centre $z_0 = \frac{1}{2}(1 + 3i)$ and radius $r = 1$. Let $z_1 = 1 + i$ and the complex number $z_2$ be outside the circle $C$ such that $|z_1 - z_0| |z_2 - z_0| = 1$. If $z_0, z_1$ and $z_2$ are collinear,then the smaller value of $|z_2|^2$ is equal to $.............$.

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

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

If $z = x + iy$ and $\omega = \frac{1 - iz}{z - i}$,then $|\omega| = 1$ shows that in the complex plane:

If $z=x+iy$, where $x$ and $y$ are real numbers and $i=\sqrt{-1}$, then the points $(x, y)$ for which $\frac{z-1}{z-i}$ is real, lie on