All the points in the set $S = \left\{ \frac{\alpha + i}{\alpha - i} : \alpha \in R \right\} (i = \sqrt{-1})$ lie on a

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

If $\frac{z-1}{z+1}$ is purely imaginary, then

$A$ circle whose radius is $r$ and centre is $z_0$,then the equation of the circle is

$A$ particle $P$ starts from the point $z_0 = 1 + 2i$,where $i = \sqrt{-1}$. It moves first horizontally away from the origin by $5$ units and then vertically away from the origin by $3$ units to reach a point $z_1$. From $z_1$,the particle moves $\sqrt{2}$ units in the direction of the vector $\hat{i} + \hat{j}$ and then it moves through an angle $\frac{\pi}{2}$ in the anticlockwise direction on a circle with the center at the origin,to reach a point $z_2$. The point $z_2$ is given by:

The number of elements in the set $\{ z = a + ib \in \mathbb{C} : a, b \in \mathbb{Z} \text{ and } 1 < |z - 3 + 2i| < 4 \}$ is: