Let $z \in \mathbb{C}$ be such that $|z| < 1$. If $w = \frac{5 + 3z}{5(1 - z)}$,then

If $z$ is a complex number,then the curves $|z|=1$,$|z-2|=1$,and $|z-1|=0$ have a common point at

The number of complex numbers $z$ such that $|z - 1| = |z + 1| = |z - i|$ is equal to

The locus of $z$ satisfying $\left|\frac{z-i}{z-2i}\right|=2$ is a

Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X = \{ z \in \mathbb{C} : \operatorname{Re}(a z^2 + bz) = a \text{ and } \operatorname{Re}(b z^2 + az) = b \}$ is equal to

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 :