I have some basics concept that, in $\mathbb{C}$, a "simply connected domain $D$" is a region in $\mathbb{C}$ with no holes. I am not sure whether it has various formal definition , but the one used in my complex analysis class looks very complicated. I barely understand what it means :
$\textbf{Definition}$ Let $\gamma_0, \gamma_1$ be two continuous closed path in a domain $D \subseteq \mathbb{C}$ parameterized by $I = [0,1]$;that is, $\gamma_i(0) = \gamma_i(1)$ for all $i = 0, 1.$ We say $\gamma_0, \gamma_1$ are $\textbf{homotopic as closed paths on D}$ if there exists function $\delta : I \times I \rightarrow D$ such that
$1) \ \delta(t,0) = \gamma_0(t) \ \forall t \in I \ \ $ $2) \ \delta(t,1) = \gamma_1(t) \ \forall t \in I \ \ $ $3) \ \delta(0,u) = \delta(1,u) \ \forall u \in I.$
A continuous closed path is $\textbf{homotopic to a point}$ if it is homotopic to a constant path(as a closed path).
$\textbf{Definition}$ A region $D \subseteq \mathbb{C}$ is called $\textbf{simply conected}$ if every continuous closed path in $D$ is homotopic to a point in $D$.
$\textbf{Problem :}$ 1) Let $D_1, D_2$ be simply connected plane domains whose intersection is nonempty and connected. Prove that their intersection and union are both simply connected.
2) Let $P, Q$ be smooth functions on a domain $D \subseteq \mathbb{C}$, Find necessary and sufficient condition for the form $P dz + Q d\bar{z}$ to be closed.
Best Answer
There is a simpler definition of simply connected for subsets of the plane. A set is simply in connected in $\Bbb C$ if, and only if its complement is connected.