The area of the triangle whose vertices are complex numbers $z, iz, z + iz$ in the Argand diagram is

Let the curve $z(1+i)+\bar{z}(1-i)=4, z \in \mathbb{C}$,divide the region $|z-3| \leq 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha-\beta|$ equals :

$A$ complex number $z$ is said to be unimodular if $|z| = 1$. Suppose $z_1$ and $z_2$ are complex numbers such that $\frac{z_1 - 2z_2}{2 - z_1 \overline{z_2}}$ is unimodular and $z_2$ is not unimodular. Then the point $z_1$ lies on a:

If the least and the largest real values of $\alpha,$ for which the equation $z+\alpha|z-1|+2i=0$ ($z \in \mathbb{C}$ and $i=\sqrt{-1}$) has a solution,are $p$ and $q$ respectively; then $4(p^2+q^2)$ is equal to ..........

Let $z_1, z_2 \in \mathbb{C}$ such that $|z_1 + z_2| = \sqrt{3}$ and $|z_1| = |z_2| = 1$. Then the value of $|z_1 - z_2|$ 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 upwards parallel to the positive $y$-axis by $3$ units to reach a point $Z_1$. From $Z_1$,the particle moves $\sqrt{2}$ units in the direction of vector $\hat{i} + \hat{j}$ and then it moves through an angle $\frac{\pi}{2}$ in the anticlockwise direction on a circle with the centre at the origin to reach point $Z_2$. Then $Z_2 =$