1. ## Rectangle Theorem

Theorem. Suppose $f$ is entire and $\Gamma$ is the boundary of a rectangle $R$. Then $\int_{\Gamma} f(z) \ dz = 0$.

Are we assuming that the rectangle bounds the curve? Or the curve bounds the rectangle?

So the general idea of the proof is that we bisect a rectangle infinitely many times? Because then we have $\int_{\Gamma} f = \sum_{i=1}^{4} \int_{\Gamma_{i}} f$. Then why is it that from some $\Gamma_k, \ 1 \leq k \leq 4$ we have $\int_{\Gamma^{(1)}} f(z) \ dz \gg I/4$?

2. Originally Posted by manjohn12
Theorem. Suppose $f$ is entire and $\Gamma$ is the boundary of a rectangle $R$. Then $\int_{\Gamma} f(z) \ dz = 0$.

Are we assuming that the rectangle bounds the curve? Or the curve bounds the rectangle?

So the general idea of the proof is that we bisect a rectangle infinitely many times? Because then we have $\int_{\Gamma} f = \sum_{i=1}^{4} \int_{\Gamma_{i}} f$. Then why is it that from some $\Gamma_k, \ 1 \leq k \leq 4$ we have $\int_{\Gamma^{(1)}} f(z) \ dz \gg I/4$?
The meaning of $\oint_{\Gamma} f(z) dz = 0$ here is that you parametrize the rectangle and then integrate over this curve.

Here "rectangle" means a figure consisting of sides made out of lines which are parallel and perpendicular to the x-y axes. "Rectangle" here does not mean the interior of the rectangle. Remember contour integration is over curves.

3. Originally Posted by ThePerfectHacker
The meaning of $\oint_{\Gamma} f(z) dz = 0$ here is that you parametrize the rectangle and then integrate over this curve.

Here "rectangle" means a figure consisting of sides made out of lines which are parallel and perpendicular to the x-y axes. "Rectangle" here does not mean the interior of the rectangle. Remember contour integration is over curves.
So suppose you have a curve $C_1$ and you take $\int_{C_1} f(z) \ dz$. We can find the boundary of a rectangle $\Gamma_1$ such that $\int_{C_1} f(z) \ dz = \int_{\Gamma_1} f(z) \ dz$? E.g. the boundary of the rectangle "contains" $C_1$?