Metamath Proof Explorer


Theorem 1cubrlem

Description: The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015)

Ref Expression
Assertion 1cubrlem ⊢ − 1 2 3 = - 1 + i ⁢ 3 2 ∧ -1 2 3 2 = - 1 - i ⁢ 3 2

Proof

Step Hyp Ref Expression
1 neg1cn ⊢ − 1 ∈ ℂ
2 neg1ne0 ⊢ − 1 ≠ 0
3 2re ⊢ 2 ∈ ℝ
4 3nn ⊢ 3 ∈ ℕ
5 nndivre ⊢ 2 ∈ ℝ ∧ 3 ∈ ℕ → 2 3 ∈ ℝ
6 3 4 5 mp2an ⊢ 2 3 ∈ ℝ
7 6 recni ⊢ 2 3 ∈ ℂ
8 cxpef ⊢ − 1 ∈ ℂ ∧ − 1 ≠ 0 ∧ 2 3 ∈ ℂ → − 1 2 3 = e 2 3 ⁢ log ⁡ -1
9 1 2 7 8 mp3an ⊢ − 1 2 3 = e 2 3 ⁢ log ⁡ -1
10 logm1 ⊢ log ⁡ -1 = i ⁢ π
11 10 oveq2i ⊢ 2 3 ⁢ log ⁡ -1 = 2 3 ⁢ i ⁢ π
12 ax-icn ⊢ i ∈ ℂ
13 pire ⊢ π ∈ ℝ
14 13 recni ⊢ π ∈ ℂ
15 7 12 14 mul12i ⊢ 2 3 ⁢ i ⁢ π = i ⁢ 2 3 ⁢ π
16 11 15 eqtri ⊢ 2 3 ⁢ log ⁡ -1 = i ⁢ 2 3 ⁢ π
17 16 fveq2i ⊢ e 2 3 ⁢ log ⁡ -1 = e i ⁢ 2 3 ⁢ π
18 6nn ⊢ 6 ∈ ℕ
19 nndivre ⊢ π ∈ ℝ ∧ 6 ∈ ℕ → π 6 ∈ ℝ
20 13 18 19 mp2an ⊢ π 6 ∈ ℝ
21 20 recni ⊢ π 6 ∈ ℂ
22 coshalfpip ⊢ π 6 ∈ ℂ → cos ⁡ π 2 + π 6 = − sin ⁡ π 6
23 21 22 ax-mp ⊢ cos ⁡ π 2 + π 6 = − sin ⁡ π 6
24 2cn ⊢ 2 ∈ ℂ
25 2ne0 ⊢ 2 ≠ 0
26 divrec2 ⊢ π ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → π 2 = 1 2 ⁢ π
27 14 24 25 26 mp3an ⊢ π 2 = 1 2 ⁢ π
28 6cn ⊢ 6 ∈ ℂ
29 18 nnne0i ⊢ 6 ≠ 0
30 divrec2 ⊢ π ∈ ℂ ∧ 6 ∈ ℂ ∧ 6 ≠ 0 → π 6 = 1 6 ⁢ π
31 14 28 29 30 mp3an ⊢ π 6 = 1 6 ⁢ π
32 27 31 oveq12i ⊢ π 2 + π 6 = 1 2 ⁢ π + 1 6 ⁢ π
33 24 25 reccli ⊢ 1 2 ∈ ℂ
34 28 29 reccli ⊢ 1 6 ∈ ℂ
35 33 34 14 adddiri ⊢ 1 2 + 1 6 ⁢ π = 1 2 ⁢ π + 1 6 ⁢ π
36 halfpm6th ⊢ 1 2 − 1 6 = 1 3 ∧ 1 2 + 1 6 = 2 3
37 36 simpri ⊢ 1 2 + 1 6 = 2 3
38 37 oveq1i ⊢ 1 2 + 1 6 ⁢ π = 2 3 ⁢ π
39 32 35 38 3eqtr2i ⊢ π 2 + π 6 = 2 3 ⁢ π
40 39 fveq2i ⊢ cos ⁡ π 2 + π 6 = cos ⁡ 2 3 ⁢ π
41 sincos6thpi ⊢ sin ⁡ π 6 = 1 2 ∧ cos ⁡ π 6 = 3 2
42 41 simpli ⊢ sin ⁡ π 6 = 1 2
43 42 negeqi ⊢ − sin ⁡ π 6 = − 1 2
44 ax-1cn ⊢ 1 ∈ ℂ
45 divneg ⊢ 1 ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → − 1 2 = − 1 2
46 44 24 25 45 mp3an ⊢ − 1 2 = − 1 2
47 43 46 eqtri ⊢ − sin ⁡ π 6 = − 1 2
48 23 40 47 3eqtr3i ⊢ cos ⁡ 2 3 ⁢ π = − 1 2
49 sinhalfpip ⊢ π 6 ∈ ℂ → sin ⁡ π 2 + π 6 = cos ⁡ π 6
50 21 49 ax-mp ⊢ sin ⁡ π 2 + π 6 = cos ⁡ π 6
51 39 fveq2i ⊢ sin ⁡ π 2 + π 6 = sin ⁡ 2 3 ⁢ π
52 41 simpri ⊢ cos ⁡ π 6 = 3 2
53 50 51 52 3eqtr3i ⊢ sin ⁡ 2 3 ⁢ π = 3 2
54 53 oveq2i ⊢ i ⁢ sin ⁡ 2 3 ⁢ π = i ⁢ 3 2
55 3re ⊢ 3 ∈ ℝ
56 3nn0 ⊢ 3 ∈ ℕ 0
57 56 nn0ge0i ⊢ 0 ≤ 3
58 resqrtcl ⊢ 3 ∈ ℝ ∧ 0 ≤ 3 → 3 ∈ ℝ
59 55 57 58 mp2an ⊢ 3 ∈ ℝ
60 59 recni ⊢ 3 ∈ ℂ
61 12 60 24 25 divassi ⊢ i ⁢ 3 2 = i ⁢ 3 2
62 54 61 eqtr4i ⊢ i ⁢ sin ⁡ 2 3 ⁢ π = i ⁢ 3 2
63 48 62 oveq12i ⊢ cos ⁡ 2 3 ⁢ π + i ⁢ sin ⁡ 2 3 ⁢ π = − 1 2 + i ⁢ 3 2
64 7 14 mulcli ⊢ 2 3 ⁢ π ∈ ℂ
65 efival ⊢ 2 3 ⁢ π ∈ ℂ → e i ⁢ 2 3 ⁢ π = cos ⁡ 2 3 ⁢ π + i ⁢ sin ⁡ 2 3 ⁢ π
66 64 65 ax-mp ⊢ e i ⁢ 2 3 ⁢ π = cos ⁡ 2 3 ⁢ π + i ⁢ sin ⁡ 2 3 ⁢ π
67 12 60 mulcli ⊢ i ⁢ 3 ∈ ℂ
68 1 67 24 25 divdiri ⊢ - 1 + i ⁢ 3 2 = − 1 2 + i ⁢ 3 2
69 63 66 68 3eqtr4i ⊢ e i ⁢ 2 3 ⁢ π = - 1 + i ⁢ 3 2
70 9 17 69 3eqtri ⊢ − 1 2 3 = - 1 + i ⁢ 3 2
71 1z ⊢ 1 ∈ ℤ
72 root1cj ⊢ 3 ∈ ℕ ∧ 1 ∈ ℤ → -1 2 3 1 ‾ = -1 2 3 3 − 1
73 4 71 72 mp2an ⊢ -1 2 3 1 ‾ = -1 2 3 3 − 1
74 cxpcl ⊢ − 1 ∈ ℂ ∧ 2 3 ∈ ℂ → − 1 2 3 ∈ ℂ
75 1 7 74 mp2an ⊢ − 1 2 3 ∈ ℂ
76 exp1 ⊢ − 1 2 3 ∈ ℂ → -1 2 3 1 = − 1 2 3
77 75 76 ax-mp ⊢ -1 2 3 1 = − 1 2 3
78 77 70 eqtri ⊢ -1 2 3 1 = - 1 + i ⁢ 3 2
79 78 fveq2i ⊢ -1 2 3 1 ‾ = - 1 + i ⁢ 3 2 ‾
80 1 67 addcli ⊢ - 1 + i ⁢ 3 ∈ ℂ
81 80 24 cjdivi ⊢ 2 ≠ 0 → - 1 + i ⁢ 3 2 ‾ = - 1 + i ⁢ 3 ‾ 2 ‾
82 25 81 ax-mp ⊢ - 1 + i ⁢ 3 2 ‾ = - 1 + i ⁢ 3 ‾ 2 ‾
83 1 67 cjaddi ⊢ - 1 + i ⁢ 3 ‾ = − 1 ‾ + i ⁢ 3 ‾
84 neg1rr ⊢ − 1 ∈ ℝ
85 cjre ⊢ − 1 ∈ ℝ → − 1 ‾ = − 1
86 84 85 ax-mp ⊢ − 1 ‾ = − 1
87 12 60 cjmuli ⊢ i ⁢ 3 ‾ = i ‾ ⁢ 3 ‾
88 cji ⊢ i ‾ = − i
89 cjre ⊢ 3 ∈ ℝ → 3 ‾ = 3
90 59 89 ax-mp ⊢ 3 ‾ = 3
91 88 90 oveq12i ⊢ i ‾ ⁢ 3 ‾ = − i ⁢ 3
92 12 60 mulneg1i ⊢ − i ⁢ 3 = − i ⁢ 3
93 87 91 92 3eqtri ⊢ i ⁢ 3 ‾ = − i ⁢ 3
94 86 93 oveq12i ⊢ − 1 ‾ + i ⁢ 3 ‾ = - 1 + − i ⁢ 3
95 1 67 negsubi ⊢ - 1 + − i ⁢ 3 = - 1 - i ⁢ 3
96 83 94 95 3eqtri ⊢ - 1 + i ⁢ 3 ‾ = - 1 - i ⁢ 3
97 cjre ⊢ 2 ∈ ℝ → 2 ‾ = 2
98 3 97 ax-mp ⊢ 2 ‾ = 2
99 96 98 oveq12i ⊢ - 1 + i ⁢ 3 ‾ 2 ‾ = - 1 - i ⁢ 3 2
100 79 82 99 3eqtri ⊢ -1 2 3 1 ‾ = - 1 - i ⁢ 3 2
101 3m1e2 ⊢ 3 − 1 = 2
102 101 oveq2i ⊢ -1 2 3 3 − 1 = -1 2 3 2
103 73 100 102 3eqtr3ri ⊢ -1 2 3 2 = - 1 - i ⁢ 3 2
104 70 103 pm3.2i ⊢ − 1 2 3 = - 1 + i ⁢ 3 2 ∧ -1 2 3 2 = - 1 - i ⁢ 3 2