$f(z) = (1-{1 \over {z^2}}) e^{z+ {1 \over z}} $ . Find the $\int_{\vert z \vert =r} f(z) dz $
If the primitive $F(z) = e^{z+{1 \over z}}$ exists, the contour integer's value would be $0$.
But Since the $f$ is analytic on $D = \{ z \Vert 0< \vert z \vert <r \}$(Not simply connected), I can't sure the existence primitive $F(z)$ of the $f$ on $D$. If the primitive of the $f$ exists or not, would you please tell me reason why you conclude like that?
Thanks.
$F $ is a primitive of $f $ on $\mathbb C \setminus \{0\}.$
Hence the integral in question $=0. $