If $x + iy = \sqrt{\phi + i\psi}$,where $i = \sqrt{-1}$ and $\phi$ and $\psi$ are non-zero real parameters,then $\phi = \text{constant}$ and $\psi = \text{constant}$ represent two systems of rectangular hyperbolas which intersect at an angle of:

Let $A, B, C$ be three sets of complex numbers as defined below:
$A = \{z : \operatorname{Im}(z) \geq 1\}$
$B = \{z : |z - 2 - i| = 3\}$
$C = \{z : \operatorname{Re}((1 - i)z) = \sqrt{2}\}$
$1.$ The number of elements in the set $A \cap B \cap C$ is:
$(A) 0, (B) 1, (C) 2, (D) \infty$
$2.$ Let $z$ be any point in $A \cap B \cap C$. Then,$|z + 1 - i|^2 + |z - 5 - i|^2$ lies between:
$(A) 25 \text{ and } 29, (B) 30 \text{ and } 34, (C) 35 \text{ and } 39, (D) 40 \text{ and } 44$
$3.$ Let $z$ be any point in $A \cap B \cap C$ and let $w$ be any point satisfying $|w - 2 - i| < 3$. Then,$|z| - |w| + 3$ lies between:
$(A) -6 \text{ and } 3, (B) -3 \text{ and } 6, (C) -6 \text{ and } 6, (D) -3 \text{ and } 9$

If a complex number $z$ is such that $\frac{z-2i}{z-2}$ is a purely imaginary number and the locus of $z$ is a closed curve,then the area of the region bounded by that closed curve and lying in the first quadrant is

Let $z_{1}$ and $z_{2}$ be the roots of the equation $z^{2} + az + 12 = 0$. If $z_{1}$,$z_{2}$,and the origin form an equilateral triangle in the complex plane,then the value of $|a|$ is:

If $w = \frac{z}{z - \frac{1}{3}i}$ and $|w| = 1$,where $i = \sqrt{-1}$,then $z$ lies on

The points $z_1, z_2, z_3, z_4$ in the complex plane are the vertices of a parallelogram taken in order,if and only if