What's wrong with this "proof" of a too strong version of the Cauchy integral theorem?

Question

Consider the path $\gamma(t)=z=e^{it}$, $t\in[0,2\pi]$, so that $dz=ie^{it}$, and suppose $F'(z)=f(z)$ on $\gamma$.

Then $$ \int_{\gamma} f(z) dz = \int_0^{2\pi} f(e^{it}) ie^{it}dt = F(e^{2\pi i})ie^{2\pi i}-F(0)ie^{i0} = F(0)i-F(0)i = 0 $$

Further, is there a proof of the Cauchy integral theorem that doesn't involve using infinitesimal rectangles or triangles (or rely on a theorem who's proof does), or, at the very least, a way of sticking all the stuff involving infinitesimal triangles or rectangles into one theorem, so that they never need to be used again in complex analysis?

Answer 1

As for the second part, this is quoted from Wikipedia:

If one assumes that the partial derivatives of a holomorphic function are continuous, the Cauchy integral theorem can be proven as a direct consequence of Green's theorem and the fact that the real and imaginary parts of $f=u+iv$ must satisfy the Cauchy–Riemann equations in the region bounded by $\gamma$, and moreover in the open neighborhood $U$ of this region. Cauchy provided this proof, but it was later proven by Goursat without requiring techniques from vector calculus, or the continuity of partial derivatives.

We can break the integrand $f$, as well as the differential $dz$ into their real and imaginary components:

$$f=u+iv$$ $$dz=dx+i\,dy$$

In this case we have $$\oint_\gamma f(z)\,dz = \oint_\gamma (u+iv)(dx+i\,dy) = \oint_\gamma (u\,dx-v\,dy) +i\oint_\gamma (v\,dx+u\,dy)$$

By Green's theorem, we may then replace the integrals around the closed contour >$\gamma$ with an area integral throughout the domain $D$ that is enclosed by >$\gamma$ as follows: $$\oint_\gamma (u\,dx-v\,dy) = \iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right) \,dx\,dy$$ $$\oint_\gamma (v\,dx+u\,dy) = \iint_D \left( \frac{\partial u}{\partial x} -\frac{\partial v}{\partial y} \right) \,dx\,dy$$

But as the real and imaginary parts of a function holomorphic in the domain $D$, >$u$ and $v$ must satisfy the Cauchy–Riemann equations there: $$\frac{ \partial u }{ \partial x } = \frac{ \partial v }{ \partial y }$$ $$\frac{ \partial u }{ \partial y } = -\frac{ \partial v }{ \partial x }$$

We therefore find that both integrands (and hence their integrals) are zero

$$\iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right )\,dx\,dy = \iint_D \left( \frac{\partial u}{\partial y} - \frac{\partial u}{\partial y} \right ) \, dx \, dy =0$$ $$\iint_D \left( \frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} \right )\,dx\,dy = \iint_D \left( \frac{\partial u}{\partial x} - \frac{\partial u}{\partial x} \right ) \, dx \, dy = 0$$

This gives the desired result $$\oint_\gamma f(z)\,dz = 0$$