Metamath Proof Explorer


Theorem iexpcyc

Description: Taking _i to the K -th power is the same as using the K mod 4 -th power instead, by i4 . (Contributed by Mario Carneiro, 7-Jul-2014)

Ref Expression
Assertion iexpcyc ⊢ K ∈ ℤ → i K mod 4 = i K

Proof

Step Hyp Ref Expression
1 zre ⊢ K ∈ ℤ → K ∈ ℝ
2 4re ⊢ 4 ∈ ℝ
3 4pos ⊢ 0 < 4
4 2 3 elrpii ⊢ 4 ∈ ℝ +
5 modval ⊢ K ∈ ℝ ∧ 4 ∈ ℝ + → K mod 4 = K − 4 ⁢ K 4
6 1 4 5 sylancl ⊢ K ∈ ℤ → K mod 4 = K − 4 ⁢ K 4
7 6 oveq2d ⊢ K ∈ ℤ → i K mod 4 = i K − 4 ⁢ K 4
8 4z ⊢ 4 ∈ ℤ
9 4nn ⊢ 4 ∈ ℕ
10 nndivre ⊢ K ∈ ℝ ∧ 4 ∈ ℕ → K 4 ∈ ℝ
11 1 9 10 sylancl ⊢ K ∈ ℤ → K 4 ∈ ℝ
12 11 flcld ⊢ K ∈ ℤ → K 4 ∈ ℤ
13 zmulcl ⊢ 4 ∈ ℤ ∧ K 4 ∈ ℤ → 4 ⁢ K 4 ∈ ℤ
14 8 12 13 sylancr ⊢ K ∈ ℤ → 4 ⁢ K 4 ∈ ℤ
15 ax-icn ⊢ i ∈ ℂ
16 ine0 ⊢ i ≠ 0
17 expsub ⊢ i ∈ ℂ ∧ i ≠ 0 ∧ K ∈ ℤ ∧ 4 ⁢ K 4 ∈ ℤ → i K − 4 ⁢ K 4 = i K i 4 ⁢ K 4
18 15 16 17 mpanl12 ⊢ K ∈ ℤ ∧ 4 ⁢ K 4 ∈ ℤ → i K − 4 ⁢ K 4 = i K i 4 ⁢ K 4
19 14 18 mpdan ⊢ K ∈ ℤ → i K − 4 ⁢ K 4 = i K i 4 ⁢ K 4
20 expmulz ⊢ i ∈ ℂ ∧ i ≠ 0 ∧ 4 ∈ ℤ ∧ K 4 ∈ ℤ → i 4 ⁢ K 4 = i 4 K 4
21 15 16 20 mpanl12 ⊢ 4 ∈ ℤ ∧ K 4 ∈ ℤ → i 4 ⁢ K 4 = i 4 K 4
22 8 12 21 sylancr ⊢ K ∈ ℤ → i 4 ⁢ K 4 = i 4 K 4
23 i4 ⊢ i 4 = 1
24 23 oveq1i ⊢ i 4 K 4 = 1 K 4
25 1exp ⊢ K 4 ∈ ℤ → 1 K 4 = 1
26 12 25 syl ⊢ K ∈ ℤ → 1 K 4 = 1
27 24 26 eqtrid ⊢ K ∈ ℤ → i 4 K 4 = 1
28 22 27 eqtrd ⊢ K ∈ ℤ → i 4 ⁢ K 4 = 1
29 28 oveq2d ⊢ K ∈ ℤ → i K i 4 ⁢ K 4 = i K 1
30 expclz ⊢ i ∈ ℂ ∧ i ≠ 0 ∧ K ∈ ℤ → i K ∈ ℂ
31 15 16 30 mp3an12 ⊢ K ∈ ℤ → i K ∈ ℂ
32 31 div1d ⊢ K ∈ ℤ → i K 1 = i K
33 29 32 eqtrd ⊢ K ∈ ℤ → i K i 4 ⁢ K 4 = i K
34 19 33 eqtrd ⊢ K ∈ ℤ → i K − 4 ⁢ K 4 = i K
35 7 34 eqtrd ⊢ K ∈ ℤ → i K mod 4 = i K