Integrate this.
$\int_{\vert z \vert =1}(1-{1 \over {z^2}}) e^{z+ {1 \over z}} $ for $f(z) = (1-{1 \over {z^2}}) e^{z+ {1 \over z}}$
First) My colleague claim that for the closed curve $\vert z \vert =1$ and exists antiderivative of $f(z)$, Hence answer is $0$. But I'm not sure why does have a antiderivative.
Second) Plus, Are there any determine method for the existence of the antiderivative by intuitively method?
e.g.) Like the thought that antiderivative of $g(z)$ = $1 \over z$ multi-valued function it doesn't have a unique antiderivate So we can't say that there is a antiderivate of the $g(z)$. But this thought just only applied for the $1 \over z$ not applied the case $f(z)$ etc.
By inspection $f(z)$ is derivative of $e^{z+\frac 1 z}$. Hence the integral is $0$.
Existence of anti-derivative is not always obvious. No general procedure can be prescribed.