This notation comes in handy for some path integrals, but I don't know yet how this is calculated. Is it simply a change of coordinates?
Is this correct: $$z=r(t)(\cos t+i\sin t) \quad t\in\mathbb{R}/2\pi. \\ \Rightarrow dz=d[r(t)](\cos t + i\sin t)+r(t)[d(\cos t + i\sin t)] \\=[r'(t)(\cos t + i\sin t)+r(t)(-\sin t+i\cos t)]\,dt?$$
Or may be better:
How do I express $dz$ in polar coordinates?
It is correct for the special case of a path that parametrises (a part of) the circle with radius $r$ around $0$ (in a particular way, it can be parametrised differently).
Generally, if you have a (piecewise) differentiable path $\gamma \colon [a,b] \to \mathbb{C}$ and a function defined and continuous (the latter can be relaxed) on the trace of $\gamma$, then the integral
$$\int_\gamma f(z)\,dz$$
is defined as
$$\int_a^b f(\gamma(t))\cdot \gamma'(t)\, dt,$$
where the latter is the ordinary Riemann- or Lebesgue-integral (or whatever integration theory you prefer) of the function $t\mapsto f(\gamma(t))\cdot \gamma'(t)$ over the interval $[a,b]$ [the function may not be defined in finitely many points, since $\gamma$ is only piecewise differentiable, but neither the Riemann nor the Lebesgue integral notice if you arbitrarily define the function in these points].
A catchy way of memorising that is to say that $dz = \gamma'(t)\,dt$ for $z = \gamma(t)$ (and viewing it as differential forms gives a formal justification, $\gamma'(t)\,dt$ is the pull-back of $dz$ by $\gamma$).