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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.

user75619
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  • Try the solution to 36898. – bbgodfrey Aug 08 '15 at 14:56
  • 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). – LLlAMnYP Aug 11 '15 at 14:08
  • Of course, I may be completely missing the intent of the author, though, if so - my apologies. – LLlAMnYP Aug 11 '15 at 14:08
  • @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. – Michael E2 Aug 11 '15 at 16:33
  • @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. – LLlAMnYP Aug 11 '15 at 16:44
  • @LLlAMnYP I agree about the dupe, but I don't think the question is clear enough to reopen (IMO). I think the OP may have forgotten a password and that my be the reason for the lack of engagement - it seems to happen regularly. (OP has not "visited" since asking.) Or it really does answer their question. Or...etc. – Michael E2 Aug 11 '15 at 16:52
  • @MichaelE2 Like I said, I'll follow your suggestion and make my own Q+A (and hopefully attract smarter and more efficient As :-). It's an interesting exercise in expression-manipulation. – LLlAMnYP Aug 11 '15 at 18:46
  • Sorry for being absent from the discussion, but here I am now... The reason I didn't want to solve $z^2 = 1$ stems from the context of the problem I was solving. Without going too much into its details, I needed to find all values of $f(a)$ at some point $a$ for an arbitrary multi-valued function $w = f(z)$. Of course, they can be found by finding all solutions to $f^{-1}(w) = a$, but that involves obtaining inverse to $f(z)$, which can be quite complicated. Eventually, it turned out all $f(z)$s I encountered in the course of my problem were good enough, in terms of finding $f^{-1}(w)$. – user75619 Aug 12 '15 at 15:04
  • @MichaelE2 and everyone else. As promised, here's my take on this problem http://mathematica.stackexchange.com/questions/91469/returning-all-branches-of-a-multiple-valued-function Hopefully it'll be helpful to the OP and interesting for the community. – LLlAMnYP Aug 12 '15 at 15:28

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