Question: Given a meromorphic function $f:\mathbb C\to\mathbb C$ and a smooth curve $\gamma:[a,b]\to\Gamma\subset\mathbb C$ with $\gamma(a)\neq\gamma(b)$, I am tempted to think the fundamental theorem of calculus yields

$$\int\_a^b\underbrace{f'(\gamma(t))\gamma'(t)\,dt}\_{=df} = f(\gamma(t))\Big|\_{t=a}^b (\*)$$

independently of the chosen path. However, taking a closed integral through both $\gamma(a)$ and $\gamma(b)\neq\gamma(a)$, the residue theorem $\oint f'(\gamma(t))\gamma'(t)\,dt = 2\pi i\sum\_k\operatorname{Res}\_k(f')$ proves me wrong when the closed curve surrounds a singularity of $f'$ since the two paths connecting $\gamma(a)$ and $\gamma(b)$ would otherwise cancel out the residue. So now my question is:

What correction term needs to be added to $(\*)$? Can that be fixed at all, or did I miss something?

Answer: 
The definition of the span of a set $S$, denoted by $\operatorname{span}(S)$, is the set of *all* finite linear combinations of the elements of $S$. 

In your case of $S\_1$, we therefore have that $\operatorname{span}\{(0,1)\} = c \cdot (0,1) = (0,c)$ for scalars $c \in \mathbb{R}$. The scalar can be anything — $c = 1, c = 3.1$ or anything else as long as it is in $\mathbb{R}$. So yes, $0$ is in $S\_1$ as we can choose $c = 0$ to get the $0$ vector. So $\operatorname{span}\{(0,1)\}$ is much more than just a single point. If you choose different values for $c$ and plot them you will see that you'll end up with a line going through the origin. As a matter of fact, it will be the $y$-axis.

Now for $S\_2$, notice that $\operatorname{span}\{(0,1)\}$ already contains the point $(0,2)$ as if we were to choose $c = 2$ we get the point $2 \cdot (0,1) = (0,2)$. So the point $(0,2)$ in $S\_2 = \operatorname{span}\{(0,1), (0,2)\}$ is redundant. However, lets formally prove that $S\_1 = S\_2$:

By definition, $S\_2 = a \cdot(0,1) + b \cdot(0,2)$ for scalars $a,b \in \mathbb{R}$. Note that $$a\cdot(0,1) + b\cdot(0,2) = (0,a) + (0,2b) = (0,a + 2b) = (a + 2b) \cdot (0,1).$$

So we have again that $S\_2$ is the set of all linear combinations of $(0,1)$ and therefore, $S\_1 = S\_2$.

Can you see that again $0$ is in $S\_2$? Can you see also that in general $0$ is *always* in the span of a set?

Hope this helps!
$S\_1$ is a line, connecting the origin to the given point $[0,1]$, and $S\_2$ is the *same* line, because it already contains $[0,2]$, too.

${\rm span\,}(v)=\{\lambda\,v\,\mid\,\lambda\in\Bbb R\}$.

So, $[0,0]$ and $[0,2]$ both are in ${\rm span\,}([0,1])$.