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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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up vote 2 down vote accepted

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.

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Thanks Davide, I see where I got confused now. – Dedede Feb 14 '12 at 7:43

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