Get branch points of an equation

Question

If I have a common function f(z,a), zand aare valid and the function ftakes real values for all z, except for an interval (z1,z2)where it becomes complicated. How to define z1and z2(what will be in terms a) using Mathematica (or is this possible)? What are the limitations?

In the test example, consider the function f[z_,a_]=Sqrt[(z-a)(z-2a)]. For real zand athis takes real values, with the exception of the interval (a,2a)where it becomes imaginary. How to find this interval in Mathematica?

In general, I would like to know how one could find it mathematically for the general case. For a function with two such variables, it would probably be simple to draw a contour graph of the Riemann surface and observe the sectional views of the branches. But what if it is a multidimensional function? Is there a general approach that can be taken?

Answer 1

, , , "a". ( ) g (a, z) = 0, . g ^ 2 - (z - a) * (z - 2 * a) == 0. Groebnerbasis, ( , ).

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' , g () . , , , g ( ). , , , "z" a.

Solve[Eliminate[{grelation == 0, D[grelation, g] == 0}, g], z]

Out [481] = {{z → a}, {z → 2 a}}

Wolfram Research

Answer 2

( ) Reduce .

.

In[1]:= Reduce[Element[{a, z}, Reals] 
           && !Element[Sqrt[(z - a) (z - 2 a)], Reals], z]
Out[1]= (a < 0 && 2a < z < a) || (a > 0 && a < z < 2a)

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Answer 3

Mathematica. . , Mathematica - , Reduce, .

- ( , ?) , Mathematica :   f [x_, y_]: = Abs [Zeta [y + 0,5 + x * I]] * I

If this function is real for any x and any -0.5 <y <0 or 0 <y <0.5, then you will find a counterexample to the Riemann zeta hypothesis, and I am sure that Mathematica cannot give the correct answer.