Cauchy theorem for a rectangle. Here $\delta R$ will give the boundary of a rectangle taking positively.
This is a theorem of the book Complex Analysis An Introduction to The Theory of Analytic Function on One Variable by L. V. Ahlfors, chapter4: Complex Integration
Let $f(z)$ be analytic is the set $R'$ obtained from a rectangle $R$ by omiting a finite number of points $\zeta_j$. If it is true that $$\lim_{z\rightarrow \zeta_j}(z-\zeta_j)f(z) = 0$$ for all j, then $$\int_{\delta R}f(z)dz = 0$$
It is sufficient to prove the case for a single exceptional point $\zeta$, for evidently $R$ can be devided into smaller rectangle which contains at most one $\zeta_j$.
We divide $R$ into nine rectangles, as shown in the figure and apply

My doubt is the inequality $$\int_{\delta R_o} \frac{|dz|}{|z-\zeta|} < 8$$.
From where the inequality is coming? Please give me some clues. 
 A: A favorite trick for estimating a line integral $\int_\gamma f$ is that it is bounded by $M\cdot L(\gamma)$, where $|f(z)|\leq M$ for all $z$ in the trace of $\gamma$, and $L(\gamma)$ the length of the curve. 
Here we can use for $M$ the reciprocal of the minimum distance of $z$ from $\zeta$, that is the minimum distance from the edge of the square to its center. And the length should be easy to figure out. That's how they got this estimate.
A: Assume $\zeta$ is the center of $R_0$. Let $L_1,\, L_2,\, L_3,\, L_4$ the sides of $R_0$ and assume they have length $l$.
So,
$$\int_{\partial R_0} \frac{\mathrm d z}{|z-\zeta|} = \sum_{n=1}^4 \int_{L_n} \frac{\mathrm d z}{|z-\zeta|}.$$
Now, let's estimate $\int_{L_n} \frac{\mathrm d z}{|z-\zeta|}$. Let $a,\, b$ be the end points of $L_n$. A parametrization to $L_n$ is $\gamma:[0,1]\to \Bbb C$,
$$\begin{align*}
\gamma(t) &= a+(b-a)t\\
\gamma'(t)&= b-a.
\end{align*}$$
Now, for $z\in \partial R_0$ the least that $|z-\zeta|$ can be is $l/2$ and it happens when $z$ is the middle point of some side.
Then
$$\int_{L_n} \frac{\mathrm d z}{|z-\zeta|} \leq
\int_0^1 \frac{\left|\gamma'(t)\right|}{|\gamma(t)-\zeta|}\mathrm dt=l\int_0^1 \frac{\mathrm dt}{\frac l2}=2.$$
I think you can take it from here.
