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

$A$ function $f$ is defined on the complex numbers by $f(z) = (a + ib)z$,where $a, b \in \mathbb{R}^+$. This function has the property that the $f$-image of any point in the complex plane is equidistant from that point and the origin. If $|a + bi| = 10$ and $b^2 = \frac{p}{q}$,where $p, q \in \mathbb{Z}$ and $\text{gcd}(p, q) = 1$,then $p + q$ is:

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:

If $a, b, c$ and $u, v, w$ are complex numbers representing the vertices of two triangles such that $c = (1 - r)a + rb$ and $w = (1 - r)u + rv$,where $r$ is a complex number,then the two triangles

The roots of the cubic equation $(z + ab)^3 = a^3$,where $a \neq 0$,represent the vertices of a triangle of sides of length:

Let $O$ be the origin,the point $A$ be $z_1 = \sqrt{3} + 2\sqrt{2}i$,and the point $B(z_2)$ be such that $\sqrt{3}|z_2| = |z_1|$ and $\arg(z_2) = \arg(z_1) + \frac{\pi}{6}$. Then: