The maximum value of the modulus of $e^{z^2}$ on the set $\{z \in \mathbb{C} : 0 \leq \operatorname{Re}(z) \leq 1, 0 \leq \operatorname{Im}(z) \leq 1\}$ is

Let $a = \text{Im}\left( \frac{1 + z^2}{2iz} \right)$,where $z$ is any non-zero complex number. The set $A = \{ a : |z| = 1 \text{ and } z \ne \pm 1 \}$ is equal to

If $z$ is a complex number having the least absolute value and $|z - 2 + 2i| = 1$,then $z =$

Let $C$ be the circle in the complex plane with centre $z_0 = \frac{1}{2}(1 + 3i)$ and radius $r = 1$. Let $z_1 = 1 + i$ and the complex number $z_2$ be outside the circle $C$ such that $|z_1 - z_0| |z_2 - z_0| = 1$. If $z_0, z_1$ and $z_2$ are collinear,then the smaller value of $|z_2|^2$ is equal to $.............$.

Let $\alpha = 8 - 14i$,$A = \{ z \in \mathbb{C} : \frac{\alpha z - \bar{\alpha} \bar{z}}{z^2 - (\bar{z})^2 - 112i} = 1 \}$,and $B = \{ z \in \mathbb{C} : |z + 3i| = 4 \}$. Then $\sum_{z \in A \cap B} (\operatorname{Re}(z) - \operatorname{Im}(z))$ is equal to $...............$.

Let $S_{1}=\{z_{1} \in \mathbb{C}:|z_{1}-3|=\frac{1}{2}\}$ and $S_{2}=\{z_{2} \in \mathbb{C}:|z_{2}-|z_{2}+1||=|z_{2}+|z_{2}-1||\}$. Then,for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$,the least value of $|z_{2}-z_{1}|$ is: