Problem: Let $f(z) = \frac{z}{(z-1)(z-2)(z-3)}$, and $g(z) = \frac{z^2}{(z-1)(z-2)(z-3)}$. Does there exist complex analytic function $F(z)$ defined on the set $A = \{ z:|z|>4\}$ such that $F'(z) = f(z)$? Is the same true for $g(z)$?
Attempt: Let $w = 1/z$. Then $w \rightarrow 0$ as $z \rightarrow \infty$, so we can redefine $f(z)$ on $A$ with $f(w)$ on $B = \{w:|w|<1/4 \}$. We also note that since $f$ is rational with its poles outside the set $B$, it is holomorphic. Then $F(z) = \int_{0}^{z} f(w)dw$ is the primitive of $f$ on the set $B$.
My confusion lies with the function $g$. With the method above I don't see any trouble in defining a primitive for it, so did I miss something or is $g$ simply redundant?
Any help is appreciated!
Let $f$ be a rational function. There will be an analytic function defined outside a disc $D$ containing the poles of $f$ with $F'=f$ if and only if $\int_Cf(z)\,dz=0$ for some contour over a large circle encircling $D$.
This is a necessary condition. It is also sufficient, the "indefinite integral" $z\mapsto \int_{a_0}^z f(w)\,dw$ will be well-defined on the complement $U$ of $D$, where $a_0$ is a fixed element of $U$ and the integration is over any path in $U$ from $a_0$ to $z$. This is because any closed contour in $U$ will have the same winding numbers round poles as some multiple of $C$.
So, the answer for your first function will be yes, since for $f(z)=z/[(z-1)(z-2)(z-3)]$ the integral round a contour of radius $R$ will be $O(1/R)$.