Here is the question: Let $f$ be holomorphic on $\mathbb{C} -\{0 \}$. Let $s_{1}, s_{2} > 0$. Prove that if $c_{1}$ and $c_{2}$ are counterclockwise oriented squares of center $0$, and side length $l_{1}, l_{2},$ respectively, then
$$ \frac{1}{2\pi i} \oint_{c_1}f(w) dw = \frac{1}{2 \pi i}\oint _{c_2} f(w) dw $$. We are not assuming that the sides of these squares are parallel to the coordinate axes.
The book gives me a hint which says the following: The function f has a holomorphic anti-derivative on the upper half plane and the lower half plane. These are the planeThe thing is that I want to apply the cauchy integral formula. The problem is how to come up with the closed contour. How do you come up with any closed contour with the two squares? I have an idea on how to draw out the closed contour but I have no idea how to put it on latex. Please only give me hints. Try not to solve the problem completely. Thank you for your help!!!!
Assume $l_1 < l_2$ so that $c_1$ is the inner square. Say $d_k$ is the right upper corner of $c_k$ and let $e$ be the straight line that goes from $d_1$ to $d_2$. Now consider the curve $\gamma$ given by $c_1$ (with initial and end points given by $d_1$), followed by $e$, then followed by $-c_2$ ( that is, $c_2$ with reverse orientation) and finally followed by $-e$. You should now be able to show, using Cauchy’s thm, that $\int_\gamma f =0$ and conclude the desired result.