If a function $f$ has a removable singularity at $z_0$ then there exists an analytic function g such that $f(z)=g(z)$ for all z in some deleted neighborhood of $z_0$.
My question is: if $g$ is analytic on a simply connected domain call it $U$ and $z_0$ is in $U$. integral of $g$ around a closed curve in $U$ is zero by Cauchy's integral theorem. Is integral of $f$ is also zero around the same curve since Laurent expansion and power series expansion of both functions coincide at $z_0$. Thank you very much!
If you have two functions $f$ and $g$ and a path $\gamma\colon[a,b]\longrightarrow\Bbb C$ such that, for each $t\in[a,b]$, $f(\gamma(t))=g(\gamma(t))$, then $\int_\gamma f=\int_\gamma g$, since these integrals only depend upon the values that $f$ and $g$ take on the image of $\gamma$.
For intance, if $\gamma\colon[0,2\pi]\longrightarrow\Bbb C$ is defined by $\gamma(t)=e^{it}$, then $\int_\gamma\frac1z\,\mathrm dz=\int_\gamma\overline z\,\mathrm dz$, since $\gamma([0,2\pi])=\{z\in\Bbb C\mid|z|=1\}$, and $|z|=1\implies\overline z=\frac1z$.