In my lecture notes the following integral was computed: \begin{align*} \int_{S_3^+(0)} \frac{e^w+z}{z+2} dw. \end{align*}
There is written: In order to use the Cauchy Integral formula, which is \begin{align*} f(z) = \frac{1}{2\pi i} \int_{S_R^+(a)} \frac{f(w)}{w-z} dw, \end{align*} we define $f(w) := e^w+w$ and $z := -2$. Clearly $f$ is differentiable in $\mathbb C$ and we obtain \begin{align*} \int_{S_3^+(0)} \frac{e^w+z}{z+2} dw = 2\pi i (e^{-2}-2). \end{align*}
Could please someone explain to me what happened here? I don't understand, why one can set $z = -2$ because then the integrand would not be defined any more? Furthermore I don't understand why the definition of $f$ should fit here.
Thanks in advance.
Edit: If I would change the path $S_3^+(0)$ to $S_1^+(0) = e^{it}$, $t \in [0,2\pi]$, then it turned out by the Deformation Theorem, that the integral is zero. And that is because $S_1^+(0)$ is homotopic to a point in $\mathbb C \setminus \{-2\}$. Could someone explain to me please, why this is the case?
An integral over a closed path that does not contain any poles of an entire function (such as $e^w+w$) is zero. This is because the path is closed, so the starting and ending points where the integral is evaluated are equivalent, and thus we have
$$\int_a^a f(z)dz=F(a)-F(a)=0$$
When the path contains a pole, as with a function like $e^w+w\over w+2$ with a path like $S_3^+(0)\to|w|=3$, then it is possible to supply an infinitesimal cut along the path and provide a different (non-circular) path which cuts out the discontinuity and results in an integral that evaluates to zero. But this different integral is a transformation of the original, and the residue from the transformation results in the original taking on a non-zero value.
The function $e^w+w\over w+2$ over the path $S_1^+(0)\to |w|=1$ contains no pole, and therefore no transformation is necessary to obtain a path that cuts out the pole, and therefore it evaluates to zero.