Why are Mobius map conformal at infinity? I think I'm missing a subtlety!
So we know an analytic function, $f$, is conformal at $z$ iff $f'(z) \neq 0$
But we see that the derivative of Mobius transformation at $\infty$ is $0$. I am told that we just need to look at $\frac{1}{z}$ to see that the Mobius transformation is conformal at $\infty$. And of course looking at the Mobius transformation evaluated at $\frac{1}{z}$, and differentiating with respect to $z$ we see that at $z = 0$ the result is non-zero. Showing that the map is conformal at $\infty$.
But why are we doing this substitution? What exactly is going on here?