If $z=x+iy$ represents a point $P$ in the Argand plane, then the area of the region represented by the inequality $2 < |z-(1+i)| < 3$ is (in $\pi$)

The equation $\text{Im}\left( \frac{iz - 2}{z - i} \right) + 1 = 0$,where $z \in \mathbb{C}$ and $z \neq i$,represents a part of a circle having radius equal to

If ${\tan ^{ - 1}}(\alpha + i\beta ) = x + iy,$ then $x =$

Let $z_{1}$ be a fixed point on the circle of radius $1$ centered at the origin in the Argand plane and $z_{1} \neq \pm 1$. Consider an equilateral triangle inscribed in the circle with $z_{1}, z_{2}, z_{3}$ as the vertices. Then, $z_{1} z_{2} z_{3}$ is equal to

If $z_1, z_2, z_3$ are points in the Argand plane,then $\left| \begin{array}{ccc} z_1 & \overline{z_1} & 1 \\ z_2 & \overline{z_2} & 1 \\ z_3 & \overline{z_3} & 1 \end{array} \right| = $

If $a$ and $b$ are the least and the greatest values respectively of $|z_1+z_2|$,where $z_1=12+5i$ and $|z_2|=9$,then $a^2+b^2=$