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?