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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.

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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.

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  • $\begingroup$ Really ? It sounds more understandable, connected is the same as in general topology, right ? Anyway, my book does not provide this definition and I guess that I need to use the definition above to prove the problem. Can anyone suggest or explain something helpful ? $\endgroup$ – user117375 Feb 23 '15 at 13:29
  • $\begingroup$ Yes, the usual definition of connectedness. $\endgroup$ – Tim Raczkowski Feb 23 '15 at 13:31
  • $\begingroup$ Any suggestion for the problem ? I am now trying to find the proof show the equivalence of the above definition of simply connected domain and the one use connectedness. $\endgroup$ – user117375 Feb 23 '15 at 13:52
  • $\begingroup$ Its a little early for me $\endgroup$ – Tim Raczkowski Feb 23 '15 at 13:54
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    $\begingroup$ If and only if its complement in the sphere is connected, you mean. $\endgroup$ – user98602 Feb 23 '15 at 22:20
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Problem 1 a) : showing that $D_1 \cup D_2$ is simply connected is a corollary from Van Kampen's theorem. But it can be proved with elementary arguments. In fact, if $\alpha$ is a loop in $D_1 \cup D_2$, we can take the connected components $\alpha_1, \dots, \alpha_n$ of $\alpha \cap D_1$. These components exist because $D_1 \cap D_2$ is connected and hence arcwise connected (since open).

Because $D_1$ is simply connected, we can find an homotopy from $\alpha_i$ to $\tilde{\alpha_i}$ where $\tilde{\alpha_i}$ is a path with same endpoints as $\alpha_i$ and in $D_1 \cap D_2$.

Now, let $\beta$ be the loop obtained by removing each $\alpha_i$ by $\tilde{\alpha_i}$. $\beta$ is homotopic to $\alpha$ and $\beta \subset D_2$, and then homotopic to some constant loop.

Problem 1 b) You can use here the caracterisation of Tim Raczkowski. Then, $(D_1 \cap D_2)^c = D_1^c \cup D_2^c$, and the union of two connected set with non-empty intersection is connected.

Problem 2 : just think in $\mathbb R^2$ ( and think about Green's theorem too).

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  • $\begingroup$ For problem 2,I have $$P dz + Q d\bar{z} = (P+Q) dx + (Pi - Qi)dy. $$ Since, if $\omega = P \ dx + Q \ dy$ is a $C^1$ function, then $\omega$ is closed on a domain $D \subseteq \mathbb{C}$ iff $P_y = Q_x$ on $D$. So if $\omega = P \ dz + Q d \bar{z}$, then $\omega$ is closed on $D$ iff $$P_x - i P_y = -Q_x - iQ_y$$ or $$P_z = -Q_{\bar{z}}$$ if $P_z = \frac{1}{2}(P_x - i P_y), Q_{\bar{z}} = \frac{1}{2} (Q_x + i Q_y)$ (it is just a notation, not partial diff respect to $z$ or $\bar{z}$. Is it correct ? $\endgroup$ – user117375 Feb 25 '15 at 2:24
  • $\begingroup$ ???? Is my work okay ? Could anyone please check or comments ? If it is okay, then I can write down my work. Thank you in advanced and sorry for bothering you guys to much. $\endgroup$ – user117375 Feb 25 '15 at 3:27
  • $\begingroup$ could you please review my solution ? $\endgroup$ – user117375 Feb 26 '15 at 0:39

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