Why do we require the (whole) domain to be simply connected in Cauchy's theorem, residue theorem, etc.? Main Question
A very common form of Cauchy's integral theorem is

Let $A\subset\mathbf{C}$ be simply connected and open, $f:A\rightarrow\mathbf{C}$ analytic, $I$ a compact interval, and $\gamma:I\rightarrow A$ continuous (or whatever) with $\gamma(I)\subset A$. Then $\int_\gamma f(z)dz=0$.

I think it would also suffice (and not make the proof any harder) to assume that $\gamma$ is homotopic in $A$ to a point without requiring that $A$ is simply connected. Am I correct?
If so, why is it never (or seldom) stated this way? The same applies to the residue theorem and other related theorems of complex analysis.

Related
I would be less surprised if the following statement was completely obvious:

Let $A\subset\mathbf{C}$ open, $I$ a compact interval, and $\gamma:I\rightarrow A$ a continuous path homotopic in $A$ to a point. Then there exists $U\subset A$ open and simply connected with $\gamma(I)\subset U$.

I suspect this is true, and I have a vague idea how to prove it. But it's not a one-liner and I don't remember seeing it in any textbook. Am I missing something?
 A: In fact, I would argue Cauchy's integral theorem in the proper form is the fact the following three statements are equivalent for $\gamma$ a loop in some open set $\Omega$:


*

*The loop $\gamma$ is nullhomologous in $\Omega$, that is, it has index (or winding number) zero around every point in the complement of $\Omega$,

*The induced map $f\in \mathcal O(\Omega)\longmapsto \int_\gamma fdz\in \mathbb C$ is identically zero,

*For each $f\in \mathcal O(\Omega)$, Cauchy's formula holds for $\gamma$:
$$\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi-z}d\xi = I(\gamma,z) f(z).$$


Since all nullhomotopic loops are nullhomologous (by the homotopic invariance of the integral) you obtain your form of Cauchy's integral theorem. One argument in favour of putting Cauchy's theorem in the above form is that it can be restated in the following form: let $H_1(\Omega)$ be the space of chains in $\Omega$ (formal sums of loops) modulo nullhomologous chains, and let $H^1(\Omega)$ be the space of holomorphic functions modulo functions that are derivatives. Then there is a nondegenerate bilinear form
$$H_1(\Omega)\otimes H^1(\Omega)\to \mathbb C$$
that takes the class of a chain $\Gamma$ in $\Omega$ and the class of a holomorphic function $f$ in $\Omega$ and assigns it to $\int_\Gamma fdz \in \mathbb C$. This is a version of de Rham's theorem in this rather simple case. 
A: I suspect that the reason that the theorem is usually stated the way you wrote is this: "simply connected domain" is pretty easy to understand (or so students think), while homotopy is harder, and doesn't need to be introduced for any other purpose just yet. And since the book is going to apply this theorem to things like a disk (over and over!) or a rectangle (sometimes) or the upper half-plane (pretty often), it's easier to make a claim about a domain you'll be using over and over than about a particular curve. 
In particular, many of the "evaluate this improper integral on the real line using residues", in which you draw ever-larger semidisk, etc., use the same domain for infinitely many semi-circle-plus-diameter paths. It's conceptually far simpler to prove once that the whole domain is simply connected than to prove that each of those infinitely many  paths is null-homotopic.  (Sure, the null-homotopy is obvious, but I think you get my point)
That's only a guess --- it's hard to know why nyone chooses a particular way of presenting an idea --- but it certainly matches what I would want to do in a basic complex analysis course. 
As for your second statement... I'm not so sure I believe that's true. I'm thinking about things like the complement of the topologists's sine curve, or the complement of the Alexander horned sphere ... I know that neither of those is really a counterexample, and the fact that the interval is compact is likely to help with the original claim...but I don't see the proof yet at all. 
A: 
Let $A\subset\mathbf{C}$ open, $I$ a compact interval, and $\gamma:I\rightarrow A$ a continuous path homotopic in $A$ to a point. Then there exists $U\subset A$ open and simply connected with $\gamma(I)\subset U$.

Let's try this curve, where $A$ is the complement of $\{0\}$.

Certainly $\int_\gamma\frac{dz}{z} = 0$, where $\gamma$ is that contour.
