Showing that $\frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta)\,d\zeta = \frac{1}{2 \pi i} \oint_{\Psi_2} f(\zeta)\,d\zeta$ via Cauchy-Goursat? In the text "Function Theory of One Complex Variable" by Robert E.Greene and Steven G.Krantz I'm having trouble obtaining a proof of $\text{Proposition (1)}$ via Cauchy-Goursat may I have hints on finishing the proof including any alternative approaches to the problem ? The attempted proof can be followed from $\text{Lemma (1)}$
$\text{Proposition (1)}$
Let $f$ be holomorphic on $\mathbb{C} \setminus  \{ 0 \}.$ Let $s_1, s_2 > 0$. Prove that if $\Psi_{1}$ and $\Psi_{2}$ are counterclockwise oriented squares of center $0$ and side length $s_1$ and $s_2$, respectively, then in $(1.1)$
$$ \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta) \, d\zeta = \frac{1}{2 \pi i} \oint_{\Psi_2} f(\zeta) \, d\zeta \tag{1.1} $$
$\text{Remark}$
We are not assuming that the sides of these squares are parallel to the coordinate axis, An important hint is to note that the function $f$ has an holomorphic antiderative on $\, \{ z = x+iy : y > 0 \}.$ and $\, \{ z = x+iy : y > 0 \}.$ Use these to evaluate the integrals over the top and bottom halves of the given curves.
$\text{Lemma (1)}$
Recall from $\text{Proposition (1)}$ that are integrals are defined over a counterclockwise oriented square for the sake of our proof one should consider this contour with it's vertices being defined as $\phi$, $\omega$, $\Gamma$, $\gamma$ and finally $0$.To fully realize the ovulation of our integral we will need the Cauchy-Goursat Theorem which is fully developed in $(1.2)$
$\text{Theorem 1.2 (Cauchy Goursat)}$
Let $U$ be an open subset of C which is simply connected, let $f : U → C$ be a holomorphic function, and let ${\displaystyle \!\,\Gamma } \!\,$  be a rectifiable path in $U$ whose start point is equal to its end point. Then in $(1.2)$
$$\oint_\Gamma f(z)\,dz = 0. \tag{1.2}$$
In view of $(1.2)$ one trivially achieves the following developments for the LHS side of $(1.1)$ in $(1.3)$
\begin{align}
& \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta) \, d\zeta \\[10pt]
= {} & \frac{1}{2 \pi i} \bigg( \oint_\omega^\phi f(z)i \, dy + \oint_\omega^\Gamma f(z) \, dx + \oint_\Gamma^\gamma f(z) \, dx + \oint_\phi^\gamma f(x)i \, dy \bigg  ) = 0 \tag{1.3}
\end{align}
In view of the approach in $(1.3)$ we obtain similar conclusions in $(1.4)$
\begin{align}
& \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta) \, d\zeta \\[10pt]
= {} & \frac{1}{2 \pi i} \bigg( \oint_\omega^\phi f(z)i\,dy + \oint_\omega^\Gamma f(z) \, dx + \oint_\Gamma^\gamma f(z) \, dx + \oint_\phi^\gamma f(x)i\,dy \bigg  ) = 0 \tag{1.4}
\end{align}
Hence, in conclusion, from $(1.4)$ and $(1.3)$ one has
$$ \bigg( \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta)\,d\zeta = 0 \bigg) = \bigg( \frac{1}{2 \pi i} \oint_{\Psi_2} f(\zeta) \, d\zeta = 0 \bigg) $$
 A: I don't understand your notation
$$\int_{\text{path one}}^{\text{path two}}$$
You meant to split the square path into its edges, right? In this case you cannot apply Goursat, the region over which the function $f$ is holomorphic is not simply connected (it has one hole at $0$), hence the two integrals can be non-zero.
I don't have the book at hand, but I think it is saying the following.
Call $R_1, R_2$ the two squares (paths). Assume that they their corresponding edges parallel, and assume that $R_1$ is inside $R_2$. Split the in-between region into four trapezoids (or eight triangles if you wish), and call their contours $T_1, \ldots, T_4$. Note that
$$\int_{R_2} = \int_{R_1} + \int_{T_1} +  \int_{T_2} + \int_{T_3} + \int_{T_4}.$$
You can check that this is true since some paths on the right hand side cancel out (one time you integrate one direction, one time the opposite). Probably a drawing would explain better. Then by Goursat the integrals along the paths $T_i$ are zero, since they are contained in a region over which $f$ is holomorphic.
From this two facts you get proposition (1)
$$\int_{R_2} f \, dz= \int_{R_1} f \, dz.$$
For an example, try to compute the integral of $f(z) = 1/z$ along a square path of arbitrary edge length. You should get $2 \pi i$, which is the same result you get if you integrate it along a circle or arbitrary radious with the same orientation (in fact, if two paths are homotopic, or two closed paths are free homotopic, then the integral along a path is equal to the integral along the other).
Edit. I just saw your comment to the OP. Yes, you could apply the residue theorem, in which case the two integrals are just the residue at the origin, but I belive your book prove the residue theorem after this proposition.
