I saw in a problem book that
$$|dz| = -\frac{ir}{z}dz$$ when $|z|=r$.
So, I tried to prove it. First, I argued that $|dz|^2 = dz\cdot\overline{dz}\,$ and $\overline{dz}=d\overline{z}$. Then I calculated $d\overline{z}$ by expanding it assuming the Leibnitz property for $d$:
$$0=dr^2=d|z|^2=d(z\cdot\overline{z})=z\cdot d\overline{z}+ dz\cdot \overline{z}\implies d\overline{z} = -\frac{r^2}{z^2}dz$$ Then taking square roots from both sides gives
$$|dz| = \sqrt{-\frac{r^2}{z^2}dz^2}=\pm\frac{ir}{z}dz$$
I didn't know how to overrule the solution with $+$ sign. So, any help is appreciated about how to choose the right sign. An alternative proof is welcome too.
But that aside, I read in one of the comments by @RobertZ under this post that $|dz|$ gives the infinitesimal arc-length. If so, then it seems intuitively expected to have
$$\int_{|z|=r}d|z|=\color{red}{2\pi r}$$
which is correct (I had made a mistake previously). So, I read that $|dz|=dt$ but I found no justification for this. I'll appreciate it if someone explains why $|dz|=dt$ where $t$ denotes any parameterization of the curve $|z|=r$.
The minus sign comes from the fact that in the formula we usually have a $\frac{1}{i}$ which is precisely $-i$; a simple proof using polar coordinates: $z = re^{i\theta}$, $dz = ire^{i\theta}d\theta = i(\frac{z}{r})(rd\theta)$, so inverting and noting that $|dz| = rd\theta$ we get the formula you want.
There is no contradiction since indeed $\int_{|z|=r}d|z|={2\pi r}$ by inspection, while both $\int_{|z|=r}dz= \int_{|z|=r} d\bar{z} = 0$ by a simple computation or Cauchy (and conjugation)...