Metamath Proof Explorer

Theorem cxpexp

Description: Relate the complex power function to the integer power function. (Contributed by Mario Carneiro, 2-Aug-2014)

Ref Expression
Assertion cxpexp ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℕ0 ) → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 elnn0 ⊢ ( 𝐵 ∈ ℕ0 ↔ ( 𝐵 ∈ ℕ ∨ 𝐵 = 0 ) )
2 nncn ⊢ ( 𝐵 ∈ ℕ → 𝐵 ∈ ℂ )
3 nnne0 ⊢ ( 𝐵 ∈ ℕ → 𝐵 ≠ 0 )
4 0cxp ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 0 ↑𝑐 𝐵 ) = 0 )
5 2 3 4 syl2anc ⊢ ( 𝐵 ∈ ℕ → ( 0 ↑𝑐 𝐵 ) = 0 )
6 0exp ⊢ ( 𝐵 ∈ ℕ → ( 0 ↑ 𝐵 ) = 0 )
7 5 6 eqtr4d ⊢ ( 𝐵 ∈ ℕ → ( 0 ↑𝑐 𝐵 ) = ( 0 ↑ 𝐵 ) )
8 0cn ⊢ 0 ∈ ℂ
9 cxpval ⊢ ( ( 0 ∈ ℂ ∧ 0 ∈ ℂ ) → ( 0 ↑𝑐 0 ) = if ( 0 = 0 , if ( 0 = 0 , 1 , 0 ) , ( exp ‘ ( 0 · ( log ‘ 0 ) ) ) ) )
10 8 8 9 mp2an ⊢ ( 0 ↑𝑐 0 ) = if ( 0 = 0 , if ( 0 = 0 , 1 , 0 ) , ( exp ‘ ( 0 · ( log ‘ 0 ) ) ) )
11 eqid ⊢ 0 = 0
12 11 iftruei ⊢ if ( 0 = 0 , if ( 0 = 0 , 1 , 0 ) , ( exp ‘ ( 0 · ( log ‘ 0 ) ) ) ) = if ( 0 = 0 , 1 , 0 )
13 11 iftruei ⊢ if ( 0 = 0 , 1 , 0 ) = 1
14 10 12 13 3eqtri ⊢ ( 0 ↑𝑐 0 ) = 1
15 0exp0e1 ⊢ ( 0 ↑ 0 ) = 1
16 14 15 eqtr4i ⊢ ( 0 ↑𝑐 0 ) = ( 0 ↑ 0 )
17 oveq2 ⊢ ( 𝐵 = 0 → ( 0 ↑𝑐 𝐵 ) = ( 0 ↑𝑐 0 ) )
18 oveq2 ⊢ ( 𝐵 = 0 → ( 0 ↑ 𝐵 ) = ( 0 ↑ 0 ) )
19 16 17 18 3eqtr4a ⊢ ( 𝐵 = 0 → ( 0 ↑𝑐 𝐵 ) = ( 0 ↑ 𝐵 ) )
20 7 19 jaoi ⊢ ( ( 𝐵 ∈ ℕ ∨ 𝐵 = 0 ) → ( 0 ↑𝑐 𝐵 ) = ( 0 ↑ 𝐵 ) )
21 1 20 sylbi ⊢ ( 𝐵 ∈ ℕ0 → ( 0 ↑𝑐 𝐵 ) = ( 0 ↑ 𝐵 ) )
22 oveq1 ⊢ ( 𝐴 = 0 → ( 𝐴 ↑𝑐 𝐵 ) = ( 0 ↑𝑐 𝐵 ) )
23 oveq1 ⊢ ( 𝐴 = 0 → ( 𝐴 ↑ 𝐵 ) = ( 0 ↑ 𝐵 ) )
24 22 23 eqeq12d ⊢ ( 𝐴 = 0 → ( ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) ↔ ( 0 ↑𝑐 𝐵 ) = ( 0 ↑ 𝐵 ) ) )
25 21 24 syl5ibrcom ⊢ ( 𝐵 ∈ ℕ0 → ( 𝐴 = 0 → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) ) )
26 25 adantl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℕ0 ) → ( 𝐴 = 0 → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) ) )
27 26 imp ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℕ0 ) ∧ 𝐴 = 0 ) → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) )
28 nn0z ⊢ ( 𝐵 ∈ ℕ0 → 𝐵 ∈ ℤ )
29 cxpexpz ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℤ ) → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) )
30 29 3expa ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ 𝐵 ∈ ℤ ) → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) )
31 28 30 sylan2 ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ 𝐵 ∈ ℕ0 ) → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) )
32 31 an32s ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℕ0 ) ∧ 𝐴 ≠ 0 ) → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) )
33 27 32 pm2.61dane ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℕ0 ) → ( 𝐴 ↑𝑐 𝐵 ) = ( 𝐴 ↑ 𝐵 ) )