Branch Cuts of $f(z) + g(z) = \sqrt{p(z)} + \sqrt{q(z)}$ How does one find the branches of 

$$f(z) + g(z) = \sqrt{p(z)} + \sqrt{q(z)}$$

where $p$ &amp; $q$ are second degree polynomials?

It would be very nice to see this general method applied to, say,

$$f(z) + g(z) = \sqrt{2z^2 + 3z - 1} + \sqrt{z^2 + 4z - 2}$$

or something better that's nice though similar.

This is apparently an impossible problem, evading just about everybody in our class, though it looks so simple.
Each quadratic has two zeros.  A branch cut starts and ends at these zeros.
Take one path between the zeros of the first quadratic, and another path between the zeros of the second quadratic.
You can have curves instead of straight lines to make them avoid each other.
A path that avoids these cuts can't go around one zero without going around the quadratic's other zero.  Both square-roots change by a factor of $e^{i\pi}$ as the path goes around both zeros.  Their product changes by a full $e^{2i\pi}$.  That is why the square-root of the quadratic is well-defined on the cut plane.
Each quadratic has two zeros.  A branch cut starts and ends at these zeros.
Take one path between the zeros of the first quadratic, and another path between the zeros of the second quadratic.
You can have curves instead of straight lines to make them avoid each other.
A path that avoids these cuts can't go around one zero without going around the quadratic's other zero.  Both square-roots change by a factor of $e^{i\pi}$ as the path goes around both zeros.  Their product changes by a full $e^{2i\pi}$.  That is why the square-root of the quadratic is well-defined on the cut plane.