# Confusion about small preliminary version of Cauchy's theorem.

I reading a proof of a preliminary version of Cauchy's theorem, on page 109 of Ahlfor's Complex Analysis.

He's proving that if the function $f(z)$ is analytic on a rectangle $R$, then $\int_{\partial R}f(z)dz=0$.

His notation is that $\eta(R)=\int_{\partial R}f(z)dz$. He breaks $R$ into four congruent rectangles, so $\eta(R)=\eta(R^{(1)})+\cdots+\eta(R^{(4)})$, and chooses some rectangle, denote it $R_1$ such that $|\eta(R^{(k)}|\geq\frac{1}{4}|\eta(R)|$. Repeating this process the rectangles converge to some point $z^*$.

Then given $\epsilon$, I know you can chooise $\delta>0$ such that for $|z-z^*|<\delta$, $$\left|\frac{f(z)-f(z^*)}{z-z^*}-f'(z^*)\right|<\epsilon$$ or $$|f(z)-f(z^*)-(z-z^*)f'(z^*)|<\epsilon|z-z^*|.$$

Since $\int_{\partial R_n}dz=0$ and $\int_{\partial R_n}zdz=0$, he writes $$\eta(R_n)=\int_{\partial R_n}[f(z)-f(z^*)-(z-z^*)f'(z^*)]dz.$$

How exactly is this final equality justified? I don't see where it's coming fr4om, thanks.

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We have $\eta(R_n)=\int_{\partial R_n}f(z)dz$. Since the integral of constants on $\partial R_n=0$ we have $\int_{\partial R_n}f(z^*)dz=0=\int_{\partial R_n}z^*f'(z^*)dz$ and $\int_{\partial R_n}zf'(z^*)dz=f'(z^*)\int_{\partial R_n}zdz=0$. Using the linearity of the integral we get the wanted result.