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How do I prove Cauchy's integral formula?

Namely:

Let $D$ be a simple, connected domain in $\mathbb C$ and $C$ be a simple, closed, anti-clockwise oriented curve contained in $D$.

Let $z_0$ be a point in the interior of the domain bounded by $C$, and suppose that $f$ is holomorphic and that $f'$ is continuous on $D$. Then:

$$\int_{C} \frac{f(z)}{z-z_0} dz = 2 \pi i f(z_0)$$

I know that we have to use, somewhere, Cauchy's integral theorem. That's if $D$ is simply connected in $\mathbb C$ and $f$ is holomorphic and has a continuous derivative, and if $C$ is a simple, closed, anti-clockwise oriented curve in $D$, then $\int_{C} f(z) dz = 0$.

But the function $z \rightarrow \frac{f(z)}{z - z_0}$ isn't continuous or differentiable on $D$, so we have to decompose the domain of integration into two parts. The first part will give an integral of $0$ value, and the second part will be the one giving the result.

The first thing that came to my mind is adding $f(z_0)$ and subtracting it in the numerator of $\frac{f(z)}{z - z_0}$, because it is tempting. Afterwards, I should use the continuity or the fact that $f$ is holomorphic to do something with that. But still, I'm stuck on how I should decompose the domain of integration and start off.

Can someone give me a hand at this? Maybe just explain to me how I should decompose the domain of integration.

I know that I could have found an answer online, but I want a hint or some help with the starting point in order to work it myself.

Thanks.

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  • $\begingroup$ You want $0$ on the right, not $2\pi i f(z_0).$ $\endgroup$ – zhw. Jun 1 '15 at 19:08
  • $\begingroup$ That's a huge mistake. @zhw. thanks for pointing out, I'll edit the question. $\endgroup$ – user230734 Jun 1 '15 at 19:20
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$$ \int_C \frac{dz}{z-z_0} = 2\pi i. $$ Hence $$ \int_C \frac{f(z_0)}{z-z_0}\,dz = 2\pi i f(z_0). \tag 1 $$ If you can show that the integral in $(1)$ is the same as $$ \int_C \frac{f(z)}{z-z_0} \, dz $$ then you've got $$ \int_C\frac{f(z)}{z-z_0}\,dz = 2\pi i f(z_0), $$ which is the result.

It is enough to show that $$ \int_C \frac{f(z)-f(z_0)}{z-z_0}\,dz = 0. $$ Let $$ g(z) = \begin{cases} \dfrac{f(z)-f(z_0)}{z-z_0} & \text{if }z\ne z_0, \\[10pt] f'(z_0) & \text{if }z=z_0. \end{cases} $$ See if you can prove $g$ is holomorphic and try to get it from there.

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