For which integers is there a holomorphic function that satisfies this? I'm stuck on the following problem in my preparation for an upcoming exam: 
Determine for which integer values of $n$ (positive, negative, or $0$), there exists a holomorphic function defined in the region $|z| > 1$, whose derivative is $\frac{z^n}{1+z^2}$.
I would just appreciate a hint that can help me solve the problem. Thank you in advance
 A: If $f$ is analytic on a connected open set $U$ there exists $F$ analytic on $U$ such that $F' = f$ iff for all closed-curve $\gamma \subset U, \int_\gamma f(z)dz = 0$. 

Proof : if there is a primitive then $\int_a^b f(z)dz = F(b)-F(a)$ whatever is the path of integration $a \to b$, conversely if for all closed curve  $\int_\gamma f(z)dz = 0$ then $F(s) = \int_{s_0}^s f(z)dz$ doesn't depend on the path of integration, it is analytic and $F' = f$.

For simply connected domains $V \subset U$ we know from the Cauchy integral formula that $\gamma \subset V \implies \int_\gamma f(z)dz$. Thus it suffices to check the condition for a few curves $C_j$ generating the fundamental group of $U$. With $U = \{ |z| > 1\}$ the fundamental group is generated by the curve $C(t ) = 2 e^{it}, t\in [0,2\pi]$. Use the residue theorem to evaluate $\int_C f(z)dz$.
A: Note that$$\lvert z\rvert>1\implies\frac{z^n}{1+z^2}=-z^{n-2}+z^{n-4}-z^{n-6}+\cdots\tag1$$So, if $n=2k-1$ for some natural number $k$, one (and only one) of the exponents in this series will be $-1$ and therefore, if $\gamma(t)=2e^{it}$ ($t\in[0,2\pi]$), then $\int_\gamma\frac{z^n}{1+z^2}\,\mathrm dz=\pm2\pi i\neq0$. So, $\frac{z^n}{1+z^2}$ has no primitive.
For the other values of $n$ (that is, when $n$ is not an odd natural number), it follows from $(1)$ that $\int_\gamma\frac{z^n}{1+z^2}\,\mathrm dz=0$, and therefore $\frac{z^n}{1+z^2}$ has a primitive.
