General method: show subset of $\mathbb{C}$ is connected Consider the two sets
$$ A = \{z \in \mathbb{C} : |z^2 - 3| < 1\}, ~~~~ B = \{z \in \mathbb{C} : |z^2 - 1| < 3 \} $$
$B$ is connected, while $A$ is not. However, I have no idea how to prove this. I know if they are connected because I used WolframAlpha, but I do not know of any method that can be used to determine if $|P(z)| < a$, where $P$ is a polynomial with $\text{deg}(P) > 1$ and $a > 0$, is connected. However, if we have $|(z - 3)^2| < 1$, then we immediately see that this is the open disk with center 3 and radius one, and we know that disks are connected. But is there a strategy for more general sets? Like
$$ \{z \in \mathbb{C} : ||z|^2 - 2| < 1\}, ~~~~ \{z \in \mathbb{C} : |z^8 + 4z| < 3\} $$
 A: It does not give rise to a general method, but at least it is a first step and easily handles the cases of your sets $A$ and $B$:
First of all, note that that these sets are open, i.e. connected and path connected are the same notions in this case.
We have the following proposition:

Let $P$ be a non-constant polynomial and $a>0$. Then any connected component of
  $M := \{|P(z)| < a\}$ contains a root of $P$.

Proof: Let $Z$ be a connected component. Note that $Z$ is open in $M$ and $M$ is open in $\mathbb C$, hence $Z$ is open in $\mathbb C$. Also note that - by a standard argument regarding the behavior of growth for polynomials - $Z$ is bounded.
For any $z \in \partial Z$, we have $|P(z)|=a$. This shows: If $P$ has no root in $Z$, it has no root in $\overline Z$. $\overline Z$ is compact, hence $|P|$ admits a minimal value on $\overline Z$ and by our previous thoughts it is clearly not admitted on the boundary. Thus, if $P$ has no root in $Z$, $\frac{1}{|P|}$ does not admit its maximal value on the boundary of $\overline Z$. This contradicts the maximum principle.

As a corollary we deduce that such a set is connected if and only if all roots are contained in the same connected component, i.e. if we find some path between the roots.
In the cases of $A$ and $B$ it is easy to see that such a path does not exist for $A$, but it exists for $B$.
