Poles on the curve Say I have this integral: $$\oint_\gamma f(z)\,{\rm d}z,$$and $f$ has a pole on $\gamma$. I understand that we "cut around" the pole with an arc of radius $\epsilon$ and then make $\epsilon \to 0$. What I can't understand is if there's a difference if we go outside or inside, like:

From the answers here I understand that we'll get different values, but if we stick to the principal value, will it be the same?

If the original curve is differentiable at the bad point, and if the pole is simple, then the PV is $2\pi i$ times half of the residue.
Too  many $\gamma$s. Let's say $\gamma$ is the original contour, with the pole on the curve. Say $\gamma_\epsilon$ is the inner contour in your picture. And let's write $$\gamma_\epsilon=\gamma_\epsilon'+\gamma_\epsilon'',$$where $\gamma_\epsilon'$ is the original contour with that little bit near the pole omitted, and $\gamma_\epsilon''$ is the circular arc.
So the PV is $$\lim_{\epsilon\to0}\int_{\gamma_\epsilon'}f(z)\,dz$$by definition (assuming I finally understand the question). Cauchy's Theorem says this is $$-\lim_{\epsilon\to0}\int_{\gamma_\epsilon''}f(z)\,dz.$$
The point to the differentiability of the original curve is that the opening of the circular arc $\gamma_\epsilon''$ tends to $\pi$ as $\epsilon\to0$. So what you need is this:
Lemma. Say $f$ has a simple pole at the origin with residue $c$. Suppose $a<b$, and for $r>0$ define an arc $\gamma_r:[a,b]\to C$ by $\gamma_r(t)=re^{it}.$ Then $$\lim_{r\to0}\int_{\gamma_r}f(z)\,dz=(b-a)ic.$$
Proof. Write $f(z)=\frac cz+g(z)$, where $g$ is continuous at the origin. The integral of $g$ tends to $0$ as $r\to0$, and you simply calculate that the integral of $c/z$ is $(b-a)ic$ for every $r$.
