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I would like to obtain the sum (or some other reduction) from a set which is the result of a previous operation:

NSolve[z^5 == 1, z, Complexes]

{{z -> -0.809017 - 0.587785 I}, {z -> -0.809017 + 0.587785 I}, {z -> 
   0.309017 - 0.951057 I}, {z -> 0.309017 + 0.951057 I}, {z -> 1.}}

How do I address these elements individually and how can I form a sum (or product)?

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    $\begingroup$ sol = NSolve[z^5 == 1, z, Complexes]; {Plus @@ (z /. sol), Times @@ (z /. sol)} $\endgroup$
    – cvgmt
    Apr 11 at 7:28
  • 1
    $\begingroup$ Also Plus@@soln[[All,1,2]] and Times@@soln[[All,1,2]] $\endgroup$
    – user1066
    Apr 11 at 7:54

1 Answer 1

Reset to default
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$\begingroup$ 
solVals = Last /@ Flatten[NSolve[z^5 == 1, z, Complexes]]

solVals1 = NSolveValues[z^5 == 1, z, Complexes]

solVals == solVals1
(* True *)

 Plus @@ sol
(* 0.+0.I) 

 Times @@ sol
(1. + 0.I *)

For exact sum much better approach is

RootSum[#^5 - 1 &, # &]

(* 0 *)

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    $\begingroup$ For the product: Exp@RootSum[#^5 - 1 &, Log] gives $1$. $\endgroup$
    – Roman
    Apr 11 at 9:08
  • $\begingroup$ @Acus: Mathematica 12 does not recognize NSolveValues; is it a new feature? $\endgroup$
    – Vectorizer
    Apr 11 at 17:56
  • $\begingroup$ @Vectorizer Documentation says it was introduced in 2021, in version 12.3 $\endgroup$
    – Acus
    Apr 12 at 6:23
  • $\begingroup$ @Acus: Thanks. I am using 12.0 $\endgroup$
    – Vectorizer
    Apr 12 at 7:06

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