The complex number $z$ satisfying the equation $|z-i|=|z+1|=1$ is

The locus of $z=x+iy$, such that $\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)=0$ is

Let $w$ $(Im\, w \neq 0)$ be a complex number. Then the set of all complex numbers $z$ satisfying the equation $w - \overline{w}z = k(1 - z)$, for some real number $k$, is

Let $u = \frac{2z + i}{z - ki}$, where $z = x + iy$ and $k > 0$. If the curve represented by $\operatorname{Re}(u) + \operatorname{Im}(u) = 1$ intersects the $y$-axis at the points $P$ and $Q$ such that $PQ = 5$, then the value of $k$ is:

Let $a, b \in \mathbb{R}$ and the roots $\alpha, \beta$ of the equation $z^2+az+b=0$ be complex. If the origin,$\alpha$ and $\beta$ represent the vertices of an equilateral triangle on the Argand plane,then

If $z=x+iy$ and the point $P$ in the Argand plane represents $z$,then the locus of $z$ satisfying the equation $|z-2|+|z-2i|=4$ is