If $\operatorname{Re}\left(\frac{z-1}{2z+i}\right)=1,$ where $z=x+iy,$ then the point $(x, y)$ lies on a

Let $z$ be a complex number such that $|z-6|=5$ and $|z+2-6i|=5$. Then the value of $z^{3}+3z^{2}-15z+141$ is equal to

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

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

If $|z|=1$ and $z \neq \pm 1$, then all the points representing $\frac{z}{1-z^{2}}$ lie on;

Let $\omega = z \bar{z} + k_1 z + k_2 i z + \lambda(1 + i)$,where $k_1, k_2 \in R$. Let $\operatorname{Re}(\omega) = 0$ be the circle $C$ of radius $1$ in the first quadrant touching the line $y = 1$ and the $y$-axis. If the curve $\operatorname{Im}(\omega) = 0$ intersects $C$ at $A$ and $B$,then $30(AB)^2$ is equal to $.......$.