Evaluate $\int_{\gamma} \frac{ |dz| }{z} $ where $\gamma $ is the unit circle I need to evaluate $\int_{\gamma} \frac{ |dz| }{z} $ where $\gamma $ is the unit circle.  
Attempt: Let $\gamma(t) = e^{it} $ where $t \in [0,2 \pi]$. How can I get rid of the absolute values in the differential?
 A: The notation $\lvert dz\rvert$ denotes the arc-length measure. For a piecewise continuously differentiable curve $\alpha \colon [a,b] \to \mathbb{C}$ and a continuous function $f$ defined (at least) on the set $\alpha([a,b])$, we have by definition
$$\int_{\alpha} f(z)\,\lvert dz\rvert = \int_a^b f(\alpha(t))\cdot \lvert \alpha'(t)\rvert\,dt.$$
Note: The constraints on $\alpha$ and $f$ can be loosened, it suffices that $\alpha$ is rectifiable, and $f\circ\alpha$ Borel measurable and "not too large in absolute value". But the typical situation is that one has a piecewise continuously differentiable path and a continuous function.
If you plug in the definitions, you can directly evaluate your integral.
However, it may be worth mentioning that in the particular case of circles, $\lvert dz\rvert$ transforms in a way that complex analysis methods to evaluate the integral become applicable (if the function is holomorphic on a suitable domain). Namely, for the parametrisation $\alpha(t) = z_0 + re^{it}$ of the circle $\lvert z-z_0\rvert = r$, we have $\alpha'(t) = ire^{it} = i(\alpha(t) - z_0)$, so we obtain
$$\int_{\alpha} f(z)\,\lvert dz\rvert = \frac{r}{i}\cdot\int_{\lvert z-z_0\rvert = r} \frac{f(z)}{z-z_0}\,dz.$$
