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.


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