Why does Mathematica not seem to know that $i^x$ cannot be equal to $0$?
We can actually 'teach' Simplify
using the option TransformationFunctions
.
dropPows[Power[_?NumericQ, _] a_ == 0] := a == 0
Simplify[I^x a == 0, TransformationFunctions -> {Automatic, dropPows}]
(* a == 0 *)
Maybe this is a difference in versions (9.0.1), but when I do:
FullSimplify[2^x a == 0, x \[Element] Reals]
a == 0
Replacing 2
with I
gives the same answer
FullSimplify[I^x a == 0, x \[Element] Reals]
a == 0
Counterexample, of arguable importance:
FunctionExpand[I^x] /. x -> DirectedInfinity[I]
(* 0 *)
Which shows why @bill s's idea of restricting x
to a number might help simplify the expression. I don't know if FullSimplify
was necessary back then, but in V11 the following works:
Simplify[I^x a == 0, x ∈ Complexes]
(* a == 0 *)