showing that $\int_{\gamma}f(z)dz=0$ for every closed curve $\gamma$ 
Let $f\in Hol(\mathbb{C}\setminus \left \{ 1 \right \})$ and suppose $\int_{\left \{ |z|=4 \right \}}f(z)dz=0$.
  Show that for every closed, piecewise smooth curve $\gamma$ that doesn't pass through $1$, $$\int_{\gamma}f(z)dz=0$$

I can quite easily show that this integral is indeede $0$ for every closed simple curve, using Cauchy's theorem. But no result that I can think of generalizes that to any closed curve because the punctuated plane is not a convex set.
I though about proving that any piecewise smooth closed curve is a finite union of simple closed piecewise smooth curves, but such a proof seems tedious.
I'd be glad to hear ideas.
 A: We have
$$\int_{\gamma} f(z)\,dz = 0 \tag{$\ast$}$$
for all piecewise smooth closed curves $\gamma$ in the domain of $f$ if and only if $f$ has a primitive.
If $f$ has a primitive $F$, then we have
$$\int_{\gamma} f(z)\,dz = \int_a^b F'(\gamma(t))\cdot \gamma'(t)\,dt = (F\circ \gamma)\Bigr\rvert_a^b = 0$$
for all (piecewise smooth) closed curves $\gamma$ by the fundamental theorem of calculus.
Conversely, if $(\ast)$ holds for all closed $\gamma$, then for any fixed $z_0$ in the domain of $f$,
$$F(w) := \int_{\gamma_w} f(z)\,dz\,,$$
where $\gamma_w$ is any piecewise smooth curve from $z_0$ to $w$ in the domain of $f$, is well-defined and thus a primitive of $f$.
So here our task is to show that $f$ has a primitive on $\mathbb{C}\setminus \{1\}$. On the half-plane $\operatorname{Re} z > 1$, we define
$$F_1(w) = \int_4^w f(z)\,dz\,,$$
where the integral is over any piecewise smooth curve from $4$ to $w$ in that half-plane. On the half-plane $\operatorname{Im} z > 0$ we define
$$F_2(w) = F_1(2+i) + \int_{2+i}^w f(z)\,dz\,,$$
on $\operatorname{Re} z < 1$ we define
$$F_3(w) = F_2(i) + \int_i^w f(z)\,dz\,,$$
and on $\operatorname{Im} z < 0$ we define
$$F_4(w) = F_3(0) + \int_0^w f(z)\,dz\,.$$
All integrals are over arbitrary piecewise smooth curves connecting the indicated points in the respective half-plane. Then each of the $F_k$ is a primitive of $f$ on the respective half-plane, and we have $F_2(2+i) = F_1(2+i)$, whence $F_1$ and $F_2$ coincide on the intersection of their respective domains. Similarly for $F_2$ and $F_3$, and for $F_3$ and $F_4$. It remains to see that $F_4$ and $F_1$ coincide on the intersection of their domains, then it follows that we have a global primitive $F$ of $f$ by defining $F(w) = F_k(w)$ if $w$ lies in the half-plane where $F_k$ is defined.
But we have
\begin{align}
0 &= \int_{\lvert z\rvert = 4} f(z)\,dz \\
&= F_2(2\sqrt{2}(-1+i)) - F_2(2\sqrt{2}(1+i)) + F_3(2\sqrt{2}(-1-i)) - F_3(2\sqrt{2}(-1+i)) \\
&\quad + F_4(2\sqrt{2}(1-i)) - F_4(2\sqrt{2}(-1-i)) + F_1(2\sqrt{2}(1+i)) - F_1(2\sqrt{2}(1-i)) \\
&= F_4(2\sqrt{2}(1-i)) - F_1(2\sqrt{2}(1-i))\,,
\end{align}
so $F_4$ and $F_1$ coincide at one point of the intersection of their domains, and since that intersection is connected, they agree on the whole intersection.
