I study physics not maths, apologies if this is trivial.
I am following a textbook (Arfken and Weber, 5th), looking at the calculus of residues. We are performing the term by term integration of the Laurent series of a function with an isolated singularity at $z_0$. We get terms of the form
$$\oint a_n(z-z_0)^ndz \tag{1}$$
where we pull out the $a_n$ as a complex constant, all terms with $n\neq -1$ are zero by the integral theorem, we then look at the case of $n=-1$ and perform the change of variables $z\rightarrow re^{i\theta}$, this step has lost me
$$a_{-1}\oint (z-z_0)^{-1} dz=a_{-1}\oint \frac{ire^{i\theta} d\theta}{re^{i\theta}}=2\pi a_{-1} \tag{2}.$$
I understand that once we have the integral in this form we just cancel the $re^{i\theta}$ and get a trivial integral of $i$ around a circle, but where is the $z_0$ going in this step, it seems like we are losing it? In other words why aren't we getting
$$a_{-1}\oint \frac{ire^{i\theta}d\theta}{re^{i\theta}-r_0e^{i\theta_0}} \tag{3}$$