The inequality $|z - 4| < |z - 2|$ represents the region given by

For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda > 1$,if $\sqrt{\lambda - 1}$ is the radius of the circle $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$,then $|\alpha - \beta|$ is equal to $.............$.

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

Let the locus of a point $z$ in the Argand plane satisfying the condition $\operatorname{Re}(z^2)=4$ be $C_1$ and the locus of $z$ satisfying the condition $\operatorname{Im}(z^2)=4$ be $C_2$. Then the number of common points of the two curves $C_1$ and $C_2$ are

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

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 $a^2+b^2=$