If three complex numbers are in $A.P.$,then they lie on

The area (in sq units) of the triangle whose vertices are the points represented by the complex numbers $0, z$,and $z e^{i \alpha}$ $(0 < \alpha < \pi)$ is:

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

If $|z - 3 - 4i| = 4$,where $i = \sqrt{-1}$,then the maximum possible value of $|z|$ is:

Let $S = \{z \in \mathbb{C} - \{i, 2i\} : \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \}$. If $\alpha - \frac{13}{11}i \in S$ and $\alpha \in \mathbb{R} - \{0\}$,then $242\alpha^2$ is equal to

Given $z$ is a complex number such that $|z| < 2$,then the maximum value of $|iz + 6 - 8i|$ is equal to-