I'm trying to decide if exist an analytic function over the annulus $A = z \in\mathbb{C} ; 1 < |z| < 2$ such that the derivative is equal to $(z^2 +1)^{-1}$
My attempt is using the derivate of $\arctan (z)$ that is: $\tan^{-1}(z) =\frac{1}{2i} \log(\frac{1+iz}{1-iz}) $ and prove that I cannot define a continuous branch on the annulus
On
The function has a primitive in $|z|>1$ and you can define the primitive explicitly by $F(z)=\sum\limits_{k=1}^{\infty}(-1)^{n} \frac 1 {z^{2n-1}(2n-1)}$. Note that the series convrges uniformly on compact subsets of $\{z:|z|>1\}$ so $F$ is indeed a holomorphic function in this region. You can compute its derivative by differentiating term by term and you get $F'(z)=-\sum\limits_{k=1}^{\infty}(-1)^{n} z^{-2n}$. But the given function $f$ can we written as $f(z)=\frac 1 {z^{2}} (1+\frac 1 {z^{2}})^{-1}$ and expanding $(1+\frac 1 {z^{2}})^{-1}$ we see that $F'=f$.
Yes, there is. Take any loop $\gamma$ in that annulus. Then the winding numbers of $\gamma$ with respect to both $i$ and $-i$ are the same. Besides, $\operatorname{res}_{z=\pm i}\frac1{z^2+1}=\mp\frac i2$. Therefore, by the residue theorem, $\int_\gamma\frac{\mathrm dz}{z^2+1}=0$. Since this happens for every loop, your functions has a primitive: just fix a point $z_0\in A$, for each $w\in A$ consider a path $\eta$ going from $z_0$ to $w$ and define$$F(w)=\int_\eta\frac{\mathrm dz}{z^2+1}.$$