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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.

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marked as duplicate by MarcoB, m_goldberg, Oleksandr R., ilian, Öskå Aug 10 '15 at 20:22

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  • $\begingroup$ Try the solution to 36898. $\endgroup$ – bbgodfrey Aug 8 '15 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 '15 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 '15 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 '15 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 '15 at 16:44

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