Does $f$ have a primitive in $\{z \in \mathbb{C}: |z| >2\}$? I'm am stuck on the following problem:
Let $G=\{z \in\mathbb{C}: |z| >2\}$ and $f(z) = \frac{1}{z^4+1}$. Is there a holomorphic function on $G$ whose derivative is $f$?
I tried to prove this the following way: let $\gamma = re^{it}, 0\leq t\leq 2\pi, r>2$. Then by the residue theorem $\int_{\gamma} \frac{1}{z^4+1}dz =0$. Actually this computation holds for any any curve that encircles $\{z: |z| \leq 2\}$ by the residues at the four poles of $f$. Also, since $f$ is analytic in $G$, we know that $\int_{\gamma} f(z)dz=0$ for any closed curve containing its interior in $D$. Therefore, $\int_{\gamma}f(z)dz =0$ for all closed curves in $D$.
I came across the following theorem: An analytic function $f$ has a primitive in a region $U$ iff $\int_{\gamma}f(z)dz =0$ for all closed curves in $D$. So I guess using this theorem then I am done right?
How do I prove the theorem? Clearly if $f$ has a primitive in $U$ then $\int_{\gamma} f(z) dz =0$ but I am not sure how to prove the opposite direction.
Also, if I made errors in my reasoning above, please let me know.
 A: To prove this theorem, let $z_0\in U$ and define
$$F(z)=\int_\gamma f(z)\,dz$$
where $\gamma$ is any contour with starting point $z_0$ and finishing point $z$.
But "why is this well-defined?", I hear you ask. If $\gamma'$ is another such
contour then
$$\int_\gamma f(z)\,dz=\int_{\gamma'}f(z)\,dz$$
since the integral over the closed contour composed of $\gamma$ followed by the reverse
of $\gamma'$ is zero. Then it's easy to prove that $F'(z)=f(z)$.
Incidentally, one can avoid residue calculation to prove that the integral
$$\int_{\gamma_R}\frac{dz}{z^4+1}$$
equals zero, where $\gamma_R$ is the circle centre $0$ and radius $R>1$. This integral
is independent of $R$ and its absolute value is bounded by $2\pi R/(R^4-1)$. Letting
$R\to\infty$ gives the integral as zero.
A: In $G$ we can write
$$f(z) = \frac{1}{z^4}(1+z^{-4}) = \frac{1}{z^4}(1-z^{-4} +z^{-8}- \cdots )$$ $$ = z^{-4} -z^{-8} + z^{-12} - \cdots.$$
Does $f$ have an antiderivative in $G?$ Yes, in fact
$$F(z) = z^{-3}/(-3) - z^{-7}/(-7) + z^{-11}/11 - \cdots$$
does the job.
