$\int_{\gamma}\frac{dz}{\sqrt{1-z^2}}=2\pi$ Along the path $\gamma(t)=2e^{it}$ for $0\leq t\leq 2\pi$ Implies Sine is $2\pi$ periodic

Ok so to back up a bit, by a trig substitution we have for $f:(-1,1)\rightarrow\mathbb{R}$:

$$f(x)=\int_0^x\frac{dt}{\sqrt{1-t^2}} = \arcsin(x)$$

Now according to the notes here: Elliptic Functions Notes, the proof that for a 'large' circle centered at zero, the fact that

$$\int_{\gamma}\frac{dz}{\sqrt{1-z^2}}=2\pi$$

implies the periodicity of the sine function. Can anyone else explain to me what's going on here? I'm really lost. I mean arcsine is of course multivalued unless we choose a particular 'branch' or whatever, but I thought we chose this 'branch' implicitly when we did our trig-sub to solve the integral in the first place. I suppose by analytic continuation, this integral must be equal to $\arcsin(z)$ on $\mathbb{C}-\{-1,1\}$. But how they go from there to talking about the periodicity of sine based just off this integral's value along a closed path containing both poles I simply don't understand. Thanks.

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