# Cauchy Integral Formula Confusion

According to Cauchy's Integral Formula, we have:

Let $U$ be an open subset of the complex plane. Let $f: U \rightarrow \mathbf{C}$ be a holomorphic function. Let $\gamma$ be the boundary of some closed disk $D$ contained in $U$. Then, given some $z_0$ interior to $D$ we have

$$f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} dz$$

Now, am I making mistake in saying:

$$\frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} dz = \frac{1}{2\pi i} \int_{\gamma}(\int_{\gamma} \frac{1}{z - z_0}dz) f(z) dz$$

$$= \frac{1}{2\pi i} \int_{\gamma}2\pi if(z) dz$$

$$= \frac{1}{2\pi i} 2 \pi i \int_{\gamma}f(z) dz$$

Hence $$f(z_0) = \int_{\gamma} f(z) dz$$

And then would it not be the case that $$\int_{\gamma} f(z) dz = 0 \quad \text{and hence} \quad f(z_0) = 0$$

because $\gamma$ is a closed path?

Edit regarding first comment

Ok, I revise the above to

$$\int_{\gamma} \frac{1}{z - z_0}dz = k \int_{0}^{2 \pi} \frac{1}{e^{i \theta}}ie^{i \theta} d\theta = k2\pi i$$

where $k$ is some positive real number. This still does not change the final equality with zero however. However it is impossible that every point is always zero. What other mistake am I making?

• The first two lines of your calculation amount to $\displaystyle{\frac{1}{z-z_0}}=2\pi i$, which cannot be true for all $z\in \gamma$, so yes, there is an error. The substitution made in the first line of the calculation is not valid. – user12477 Aug 27 '12 at 22:08
• @user12477 please see edit – providence Aug 27 '12 at 22:27

If I am understanding correctly, in the first step, you are trying to apply Cauchy's Integral Formula to the function $1$ to get $$2\pi i=\oint_\gamma \frac{dz}{z-z_0}$$ The problem is you substituted $$\oint_\gamma \frac{dz}{z-z_0} = \frac{1}{z-z_0}$$ which is more or less nonsensical. If I am to guess, what you tried to substitute for is $$\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-z_0}\ dz = \frac{1}{2\pi i}\oint_\gamma \left(\frac{1}{2\pi i}\oint_\gamma \frac{1}{t-z_0}\ dt\right)\frac{f(z)}{z-z_0}\ dz$$ $$=\frac{1}{(2\pi i)^2}\oint_\gamma \oint_\gamma \frac{f(z)}{(t-z_0)(z-z_0)}\ dt\ dz$$ which is a vastly different expression.
The new substitution adds nothing new to the problem. You are still attempting the false substitution mentioned above. You mentioned that you wish to separate $f(z)$ and $\frac{1}{z-z_0}$. To do that you must have an integral expression for $g(z)=\frac{1}{z-z_0}$. However, there is no easy integral expression which represents $g$ since $g$ is not holomorphic in the region bound by $\gamma$ (it has a simple pole at $z=z_0$) and so without even mentioning the fact that the substitution you applied was incorrect, Cauchy's integral formula cannot even be applied to $g$.
• Sorry, I have made a correction above regarding the substitution I was trying to make. I still get zero coming out the end though. I don't quite follow your process. What I am trying to do is remove the numerator $f(z)$ to handle the integration of $f(z)$ and $\frac{1}{z - z_0}$ individually, if you understand my attempt at an explanation. – providence Aug 27 '12 at 22:31