Let's say I have a complex-valued function $f(z) = \sqrt{z}$. When I plug $z = 1$ into it, it returns 1. I need it to return a list $\{-1,1\}$. Is it possible to force Mathematica to do that? I know it's possible to do that by solving $z^2 = 1$, but it involves finding inverse function to $\sqrt{z}$ first. And I don't want to do that.

  • $\begingroup$ Try the solution to 36898. $\endgroup$
    – bbgodfrey
    Aug 8, 2015 at 14:56
  • $\begingroup$ I vote to reopen this question as I feel, the linked answer doesn't really address the problem at the angle, at which this question (in my interpretation) approaches it. Many functions can be multivalued, depending on their definition. If the square root is defined as the number which, when squared, gives the original value, it does have two correct values. Consider, e.g. a function f[x_, y_] := x^(1/2) + y^(1/3). What I would like to get then, is a list of six values (two possible square roots combined with each of the three possible cubic roots). $\endgroup$
    – LLlAMnYP
    Aug 11, 2015 at 14:08
  • $\begingroup$ Of course, I may be completely missing the intent of the author, though, if so - my apologies. $\endgroup$
    – LLlAMnYP
    Aug 11, 2015 at 14:08
  • $\begingroup$ @LLlAMnYP I think you've considerable expanded the expressed intent. (Why not ask and answer your own question?) I am confuzed by the OP's not wanting to solve $z^2 = 1$. If we're not allowed to use the algebraic equation defining the conjugates, I feel the problem has a very narrow scope, namely, just ±Sqrt. So the answer would simply be f[z_] := {-1, 1} Sqrt[z]. If that is not the answer, then the Q is unclear. $\endgroup$
    – Michael E2
    Aug 11, 2015 at 16:33
  • $\begingroup$ @MichaelE2 I guess I just might get round to that tomorrow. IMO, the linked question seems to tackle a much broader and far more complex problem, than just capturing every branch of "multi-valued" functions. In that respect, perhaps the answer to the OP in the link might be a bit too advanced, given the relatively simpler question. $\endgroup$
    – LLlAMnYP
    Aug 11, 2015 at 16:44


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