I'm asked to describe the Riemann surface of $f(z)=\sqrt{(z^2-1)(z^2-2)}$.
I know that the branch points are $z=\pm1,\pm2$, so if I just keep $[-2,-1]\cup[1,2]$ from the real axis, will I get the two sheets that I need to form the Riemann surface? And if so, how do I "glue" them together?
Any help will be appreciated, thanks.
Well...there are only four choices for gluing.
If you start near $z = 1.1$, then $f(z) \approx \sqrt{(z^2 -1)(1.1^2 - 2)} \approx \sqrt{(1-z^2)}$. As you walk around the circle of radius 0.1 around $z = 1$, gradually updating the $f(z)$ value continuously, when you get back to $1.1$, has the $f(z)$ value negated or is it the same as when you started? If it's negated, then the branch cut you've drawn has to be glued with a criss-croos. If it's the same, then you can sew up that branch cut.
Do the same thing at other branch points to determine the overall structure. Yeah, I'm actually suggesting you do some arithmetic here! Once you're done, you'll start to understand how you might do it without those computations, etc.
By the way...it's not clear to me that the branch cuts you mention are sufficient; there might need to be one between -1 and 1, and then two more between 2 and infinity and between -2 and -infinity.