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?

  • $\begingroup$ 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. $\endgroup$ – user12477 Aug 27 '12 at 22:08
  • $\begingroup$ @user12477 please see edit $\endgroup$ – 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.

In response to OP's edit

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 $g$.

  • $\begingroup$ 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. $\endgroup$ – providence Aug 27 '12 at 22:31
  • 1
    $\begingroup$ Ok, I realize the mistake I was making. Thank you. $\endgroup$ – providence Aug 27 '12 at 23:10

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