I have a 1-form $\omega = \frac{az+b}{z^2+1}dz$ on $U = \mathbb{C}$ \ {i,-i} and I have to find for wich $a,b \in \mathbb{C}$ $\omega$ is exact on $U$.
So I want to see if the integral along a loop is $0$ using the Cauchy´s Integral Theorem.
In order to use the Cauchy´s Integral Theorem I´ve found $c,d \in \mathbb{C}$ such that
$\omega = \frac{az+b}{z^2+1}dz = \frac{c}{z+i}dz + \frac{d}{z-i}dz$
So $c=\frac{a-bi}{2}$ and $d=\frac{a+bi}{2}$ and the expresion of the integral is the following:
$\int_{\gamma} w dz = \int_{\gamma} \frac{1}{2}\frac{a-bi}{z+i} dz + \frac{1}{2}\frac{a+bi}{z-i} w dz = (*)$
As $\gamma$ I have taken the loop around the origin of radius 2. $\gamma : [0,2\pi]\rightarrow \mathbb{C}$;
$t \rightarrow 2e^{it}$
So applying the Cauchy´s Integral Theorem we have
$(*) = \frac{1}{2}(a-bi)2\pi i + \frac{1}{2}(a+bi)2\pi i = 2a\pi i$
So w is exact on U if it is local exact and the integral around a loop is 0.
$\int_{\gamma} w dz = 0 \iff a = 0$
Now I have the following subsets of $\mathbb{C}:$
$V = \mathbb{C}$ \ { i$y$ | $-1 \leq y \leq 1$} and
$A = \mathbb{C}$ \ { i$y$ | $y \leq 1$}
And I have to see for which $a,b \in \mathbb{C}$ if $\omega\restriction A$ and $\omega\restriction V$ is exact.
I don´t see what are the consecuences of the restrictions if I apply the same argument as before.
Can anyone help me?