(Fake proof) Counterclockwise contour integral of identity function around unit circle is $-2\pi i$

Question

First, the result is obviously false by Cauchy's integral formula, given that the identity function is one of the simplest analytic functions and has no singularities. So the contour integral is zero.

Here is the outline of the fake proof. Let $C$ denote the path tracing the unit circle counterclockwise exactly once.

1. The contour integral of $z \mapsto 1 / z$ along $C$ is $2 \pi i$. (Residue theorem)

2. Along $C$, $z \mapsto 1 /z $ coincides exactly with $z \mapsto (\sqrt{z \cdot \bar{z} })/ z = e^{- i \theta}$. Hence the contour integral of $z \mapsto e^{- i \theta}$ along $C$ is also $2 \pi i$.

3. The contour integral of $z \mapsto e^{i \theta} = z / ( \sqrt{ z \cdot \bar{z} })$ along $C$ is the same as the contour integral of $z \mapsto e^{-i \theta}$ along $-C$, hence $- 2 \pi i$.

4. Along $C$, the identity function $z \mapsto z$ coincides with $z \mapsto e^{i \theta}$, hence its contour integral is the same, $- 2 \pi i$.

The step I am most suspicious of is 3., although conceivably 2. or 4. could also be incorrect.

Because 3. is so alluring from a "visual" or "geometric" perspective, even if it is false ("$e^{-i \theta}$ is clockwise rotation, $e^{i \theta}$ is counterclockwise rotation, so the integral of the latter is the negative of the integral of the former"), if the problematic step is indeed 3., I would greatly appreciate an answer that can give a "visual" or "geometric" (verbal) explanation of why step 3. is wrong.

Also I realize that $e^{-i \theta}$ and $e^{i \theta}$ aren't analytic (if they were, $\bar{z}$ would be analytic but it isn't, or alternatively by analytic continuation they would equal $1/z$ and $z$ respectively), but they are still continuous, so I don't see why that would prevent steps 2. and 4. from being correct. So if one or both of those steps are wrong, some explanation of why would also be helpful.

In case the context is helpful, this is essentially a follow-up question to section IV of chapter 8 of Needham's "Visual Complex Analysis", which basically explains the residue theorem result for the special case of $z \mapsto 1/z$.

Answer 1

The contour integral of $z$ around $C$ is $$ \oint_Cz\,dz=\int_0^{2\pi}e^{i\theta}\,i\,e^{i\theta}\,d\theta=\frac12 \int_0^{2\pi}2i\,e^{2i\theta}\,d\theta=\frac12e^{2i\theta}\,\Big|_{\theta=0}^{\theta=2\pi}=0 $$ as it should.