Represent in the cartesian plane the inequality $|w-z|<1$ given $z$

Question

I know that $A=\{z\in\mathbb{C}:z^3+z^2-3z-6=0\}$.

I found out that the solutions are: $z_1=2, z_2=\frac{-3+\sqrt{3}i}{2}, z_3=\frac{-3-\sqrt{3}i}{2}$.

Now, I have to represent in the cartesian plane $B=\{w\in \mathbb{C}:|w-z|<1, \text{with z} \in A\}$

How do I do that since $w$ can be any complex number?

Answer 1

$|w-z_0|<1$ is the interior of a circle centered at $z_0$ and radius $1$. So for each $z_0 \in A$ you draw such a circle (i.e. 3 circles).

Now assuming that the defining condition for set $B$ is $|w-z|<1$ for some $z \in A$, you want the union of these three regions as your set $B$.