Metamath Proof Explorer


Theorem cxprecd

Description: Complex exponentiation of a reciprocal. (Contributed by Mario Carneiro, 30-May-2016)

Ref Expression
Hypotheses rpcxpcld.1
|- ( ph -> A e. RR+ )
cxprecd.2
|- ( ph -> B e. CC )
Assertion cxprecd
|- ( ph -> ( ( 1 / A ) ^c B ) = ( 1 / ( A ^c B ) ) )

Proof

Step Hyp Ref Expression
1 rpcxpcld.1
 |-  ( ph -> A e. RR+ )
2 cxprecd.2
 |-  ( ph -> B e. CC )
3 cxprec
 |-  ( ( A e. RR+ /\ B e. CC ) -> ( ( 1 / A ) ^c B ) = ( 1 / ( A ^c B ) ) )
4 1 2 3 syl2anc
 |-  ( ph -> ( ( 1 / A ) ^c B ) = ( 1 / ( A ^c B ) ) )