Primitive of $f(x)$ on the punctured plane where $f(x)$ is analytic Suppose $f$ is analytic on $D = C\setminus\{0\}$ i.e. the punctured at zero complex plane. Show that there exists a constant $A$ such that $f(z) - \frac{A}{z}$ has a primitive on $D$. Find the value of $A$ in terms on an integral of $f$.   
Any help on the this problem would be greatly appreciated. I think it is related to using a laurent expansion but I'm not sure how to even get started. Thanks.

Using Laurent series is a good idea. There is a theorem that says  integrating power series term-by-term is legitimate. In the series
$$f(z) = \sum_{n=-\infty}^\infty a_n z^n$$
what is the problematic term for integration? It's the power $-1$, which integrates to logarithm, which isn't a single-valued function in the punctured plane. So, set it aside: 
$$f(z)-\frac{a_{-1}}{z} = \sum_{n\ne -1}  a_n z^n$$
and integrate:
$$\int \left(f(z)-\frac{a_{-1}}{z}\right)\,dz = \sum_{n\ne -1}  \frac{a_n}{n+1} z^{n+1} \tag{1}$$
The expression on the right is the primitive  you want. 
The number $a_{-1}$ can be found by integration of $f $ along a circle, such as the unit circle: 
$$\int_{|z|=1}f(z)\,dz = 2\pi a_{-1}$$
Indeed, the powers other than $-1$ have a primitive and therefore integrate to zero over every closed curve.
