If I have a general function,f(z,a), z and a are both real, and the function f takes on real values for all z except in some interval (z1,z2), where it becomes complex. How do I determine z1 and z2 (which will be in terms of a) using Mathematica (or is this possible)? What are the limitations?
For a test example, consider the function f[z_,a_]=Sqrt[(z-a)(z-2a)]. For real z and a, this takes on real values except in the interval (a,2a), where it becomes imaginary. How do I find this interval in Mathematica?
In general, I'd like to know how one would go about finding it mathematically for a general case. For a function with just two variables like this, it'd probably be straightforward to do a contour plot of the Riemann surface and observe the branch cuts. But what if it is a multivariate function? Is there a general approach that one can take?
What you have appears to be a Riemann surface parametrized by 'a'. Consider the algebraic (or analytic) relation g(a,z)=0 that would be spawned from this branch of a parametrized Riemann surface. In this case it is simply g^2 - (z - a)*(z - 2*a) == 0. More generally it might be obtained using Groebnerbasis, as below (no guarantee this will always work without some amount of user intervention).
grelation = First[GroebnerBasis[g - Sqrt[(z - a)*(z - 2*a)], {x, a, g}]]
Out[472]= 2 a^2 - g^2 - 3 a z + z^2
A necessary condition for the branch points, as functions of the parameter 'a', is that the zero set for 'g' not give a (single valued) function in a neighborhood of such points. This in turn means that the partial derivative of this relation with respect to g vanishes (this is from the implicit function theorem of multivariable calculus). So we find where grelation and its derivative both vanish, and solve for 'z' as a function of 'a'.
Solve[Eliminate[{grelation == 0, D[grelation, g] == 0}, g], z]
Out[481]= {{z -> a}, {z -> 2 a}}
Daniel Lichtblau Wolfram Research