# Checking branching at $\infty$ for Riemann surface given by analytic continuation

I'm looking at the ramified riemann surfaces of some algebraic functions. Say we are interested in the Riemann surface $$M_1$$ corresponding to $$w=\sqrt{(z-a)}$$ and the Riemann surface $$M_2$$ corresponding to $$w=\sqrt{(z-a)(z-b)}$$.

For the two-sheet Riemann surface $$M_1$$, if we want to find the branch points, we look at where the implicit function theorem fails for $$P(z,w)=w^2-z+a$$, i.e. we look at solutions to $$P(z,w)=0$$ and $$\partial_wP(z,w)=0$$. I.e. $$w^2-z+a=0,\qquad 2w=0,$$ in which case $$w=0$$ and $$z=a$$, i.e. there is one branch point (that isn't $$\infty$$) at $$(z,w)=(a,0)$$.

For the two-sheet Riemann surface $$M_2$$, we repeat this, where now $$w^2=(z-a)(z-b)=z^2-(a+b)z+ab$$ so that we have to solve: $$w^2-z^2+(a+b)z-ab=0,\qquad 2w=0,$$ which gives $$w=0$$ with $$z=a,b$$, i.e. we have two branch points $$\{(a,0),(b,0)\}$$ (that aren't $$\infty$$), both of order $$1$$.

How do I check if $$\infty$$ is a branch point? I thought I could consider if there are branch points at $$z=0$$ for $$P(1/z,w)=w^2-(1/z)+a=w^2z-1+az=0,$$ $$P_w(1/z,w)=2w$$ for $$M_1$$, but then $$z=0$$ doesn't arise as a branch point, and I'm told that $$z=\infty$$ is indeed a branch point (which I guess I can see must be true since the genus of this Riemann surface is $$1-2+\frac12 + \frac12B(\infty)=-\frac12+B(\infty)/2$$, i.e. $$0$$ if there is a branch point at $$z=\infty$$ of order $$1$$.)

I'm also told that $$M_2$$ has no branching at $$z=\infty$$, for comparison.