$\sqrt[n]{1-z^2}$ =$\sqrt[n]{(1-z)(1+z)}$ two branch points at -1,and 1.
$\sqrt[n]{z}$ hast n branches and only one branch point at z=0.
My understanding:
When we want to integrate on different part of riemann surface , and there is only one branch point at the origin, we can multiply the input varible z by $e^{ic}$,where c is a constant that depedns on how much we want to move along the riemann surface. And then we can set a branch cut to cut out the branch to avoid multivalued output ocurring. we multiply z by $e^{i2\pi}$ and put a branch cut along the positive real axie if we want to move to the next branch. We seperate out th single-valued part of the function.
However, if there are more than two branch points, and the branch points are not at the origin, what can we do to move along riemann surface and change the domain? I know that as soon as we encircle any branch point, multivalued output occurs, and we can consider that we have moved onto another branch. The possition of Branch cuts is somewhat abitary as long as they are set in a way that can prevent the path of integration from encircling any branch point.
Question(1): If the origin is our brnach point, multiplying z by $e^{in2\pi}$ will rotate any input for n times around the origin.Thus, the domain is moved along the riemann surface . However, if the branch point is not at 0, but at a point (a,b), what can we do algebracially to the input space to make everying we put in rotated around the point (a,b) ?
question(2): When there are two branch points Q and P, if I first encircle Q and then also encircle P, will I end up on the same part of the riemann surface surface as I will if I first encircle P and then Q? Can we algebrically do this? if we encircle both branch points all at once, where should I end up?