Metamath Proof Explorer


Theorem cos9thpiminplylem5

Description: The constructed complex number A is a root of the polynomial ( ( X ^ 3 ) + ( ( -u 3 x. X ) + 1 ) ) . (Contributed by Thierry Arnoux, 14-Nov-2025)

Ref Expression
Hypotheses cos9thpiminplylem3.1 ⊢ O = e i ⁢ 2 ⁢ π 3
cos9thpiminplylem4.2 ⊢ Z = O 1 3
cos9thpiminplylem5.3 ⊢ A = Z + 1 Z
Assertion cos9thpiminplylem5 ⊢ A 3 + -3 ⁢ A + 1 = 0

Proof

Step Hyp Ref Expression
1 cos9thpiminplylem3.1 ⊢ O = e i ⁢ 2 ⁢ π 3
2 cos9thpiminplylem4.2 ⊢ Z = O 1 3
3 cos9thpiminplylem5.3 ⊢ A = Z + 1 Z
4 ax-icn ⊢ i ∈ ℂ
5 2cn ⊢ 2 ∈ ℂ
6 picn ⊢ π ∈ ℂ
7 5 6 mulcli ⊢ 2 ⁢ π ∈ ℂ
8 4 7 mulcli ⊢ i ⁢ 2 ⁢ π ∈ ℂ
9 3cn ⊢ 3 ∈ ℂ
10 3ne0 ⊢ 3 ≠ 0
11 8 9 10 divcli ⊢ i ⁢ 2 ⁢ π 3 ∈ ℂ
12 efcl ⊢ i ⁢ 2 ⁢ π 3 ∈ ℂ → e i ⁢ 2 ⁢ π 3 ∈ ℂ
13 11 12 ax-mp ⊢ e i ⁢ 2 ⁢ π 3 ∈ ℂ
14 1 13 eqeltri ⊢ O ∈ ℂ
15 ax-1cn ⊢ 1 ∈ ℂ
16 15 9 10 divcli ⊢ 1 3 ∈ ℂ
17 cxpcl ⊢ O ∈ ℂ ∧ 1 3 ∈ ℂ → O 1 3 ∈ ℂ
18 14 16 17 mp2an ⊢ O 1 3 ∈ ℂ
19 2 18 eqeltri ⊢ Z ∈ ℂ
20 efne0 ⊢ i ⁢ 2 ⁢ π 3 ∈ ℂ → e i ⁢ 2 ⁢ π 3 ≠ 0
21 11 20 ax-mp ⊢ e i ⁢ 2 ⁢ π 3 ≠ 0
22 1 21 eqnetri ⊢ O ≠ 0
23 cxpne0 ⊢ O ∈ ℂ ∧ O ≠ 0 ∧ 1 3 ∈ ℂ → O 1 3 ≠ 0
24 14 22 16 23 mp3an ⊢ O 1 3 ≠ 0
25 2 24 eqnetri ⊢ Z ≠ 0
26 15 19 25 divcli ⊢ 1 Z ∈ ℂ
27 19 26 addcli ⊢ Z + 1 Z ∈ ℂ
28 3 27 eqeltri ⊢ A ∈ ℂ
29 3nn0 ⊢ 3 ∈ ℕ 0
30 expcl ⊢ A ∈ ℂ ∧ 3 ∈ ℕ 0 → A 3 ∈ ℂ
31 28 29 30 mp2an ⊢ A 3 ∈ ℂ
32 9 negcli ⊢ − 3 ∈ ℂ
33 32 28 mulcli ⊢ -3 ⁢ A ∈ ℂ
34 33 15 addcli ⊢ -3 ⁢ A + 1 ∈ ℂ
35 31 34 pm3.2i ⊢ A 3 ∈ ℂ ∧ -3 ⁢ A + 1 ∈ ℂ
36 binom3 ⊢ Z ∈ ℂ ∧ 1 Z ∈ ℂ → Z + 1 Z 3 = Z 3 + 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2 + 1 Z 3
37 19 26 36 mp2an ⊢ Z + 1 Z 3 = Z 3 + 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2 + 1 Z 3
38 3 oveq1i ⊢ A 3 = Z + 1 Z 3
39 33 15 negdii ⊢ − -3 ⁢ A + 1 = - -3 ⁢ A + -1
40 32 28 mulneg1i ⊢ − -3 ⁢ A = − -3 ⁢ A
41 40 oveq1i ⊢ − -3 ⁢ A + -1 = - -3 ⁢ A + -1
42 9 negnegi ⊢ − -3 = 3
43 42 oveq1i ⊢ − -3 ⁢ A = 3 ⁢ A
44 43 oveq1i ⊢ − -3 ⁢ A + -1 = 3 ⁢ A + -1
45 41 44 eqtr3i ⊢ - -3 ⁢ A + -1 = 3 ⁢ A + -1
46 6nn0 ⊢ 6 ∈ ℕ 0
47 expcl ⊢ Z ∈ ℂ ∧ 6 ∈ ℕ 0 → Z 6 ∈ ℂ
48 19 46 47 mp2an ⊢ Z 6 ∈ ℂ
49 expcl ⊢ Z ∈ ℂ ∧ 3 ∈ ℕ 0 → Z 3 ∈ ℂ
50 19 29 49 mp2an ⊢ Z 3 ∈ ℂ
51 48 50 addcomi ⊢ Z 6 + Z 3 = Z 3 + Z 6
52 1 2 cos9thpiminplylem4 ⊢ Z 6 + Z 3 = − 1
53 13 sqcli ⊢ e i ⁢ 2 ⁢ π 3 2 ∈ ℂ
54 13 21 pm3.2i ⊢ e i ⁢ 2 ⁢ π 3 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 ≠ 0
55 15 53 54 3pm3.2i ⊢ 1 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 2 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 ≠ 0
56 5 15 11 adddiri ⊢ 2 + 1 ⁢ i ⁢ 2 ⁢ π 3 = 2 ⁢ i ⁢ 2 ⁢ π 3 + 1 ⁢ i ⁢ 2 ⁢ π 3
57 2p1e3 ⊢ 2 + 1 = 3
58 57 oveq1i ⊢ 2 + 1 ⁢ i ⁢ 2 ⁢ π 3 = 3 ⁢ i ⁢ 2 ⁢ π 3
59 8 9 10 divcan2i ⊢ 3 ⁢ i ⁢ 2 ⁢ π 3 = i ⁢ 2 ⁢ π
60 58 59 eqtri ⊢ 2 + 1 ⁢ i ⁢ 2 ⁢ π 3 = i ⁢ 2 ⁢ π
61 11 mullidi ⊢ 1 ⁢ i ⁢ 2 ⁢ π 3 = i ⁢ 2 ⁢ π 3
62 61 oveq2i ⊢ 2 ⁢ i ⁢ 2 ⁢ π 3 + 1 ⁢ i ⁢ 2 ⁢ π 3 = 2 ⁢ i ⁢ 2 ⁢ π 3 + i ⁢ 2 ⁢ π 3
63 56 60 62 3eqtr3ri ⊢ 2 ⁢ i ⁢ 2 ⁢ π 3 + i ⁢ 2 ⁢ π 3 = i ⁢ 2 ⁢ π
64 63 fveq2i ⊢ e 2 ⁢ i ⁢ 2 ⁢ π 3 + i ⁢ 2 ⁢ π 3 = e i ⁢ 2 ⁢ π
65 5 11 mulcli ⊢ 2 ⁢ i ⁢ 2 ⁢ π 3 ∈ ℂ
66 efadd ⊢ 2 ⁢ i ⁢ 2 ⁢ π 3 ∈ ℂ ∧ i ⁢ 2 ⁢ π 3 ∈ ℂ → e 2 ⁢ i ⁢ 2 ⁢ π 3 + i ⁢ 2 ⁢ π 3 = e 2 ⁢ i ⁢ 2 ⁢ π 3 ⁢ e i ⁢ 2 ⁢ π 3
67 65 11 66 mp2an ⊢ e 2 ⁢ i ⁢ 2 ⁢ π 3 + i ⁢ 2 ⁢ π 3 = e 2 ⁢ i ⁢ 2 ⁢ π 3 ⁢ e i ⁢ 2 ⁢ π 3
68 64 67 eqtr3i ⊢ e i ⁢ 2 ⁢ π = e 2 ⁢ i ⁢ 2 ⁢ π 3 ⁢ e i ⁢ 2 ⁢ π 3
69 ef2pi ⊢ e i ⁢ 2 ⁢ π = 1
70 2z ⊢ 2 ∈ ℤ
71 efexp ⊢ i ⁢ 2 ⁢ π 3 ∈ ℂ ∧ 2 ∈ ℤ → e 2 ⁢ i ⁢ 2 ⁢ π 3 = e i ⁢ 2 ⁢ π 3 2
72 11 70 71 mp2an ⊢ e 2 ⁢ i ⁢ 2 ⁢ π 3 = e i ⁢ 2 ⁢ π 3 2
73 72 oveq1i ⊢ e 2 ⁢ i ⁢ 2 ⁢ π 3 ⁢ e i ⁢ 2 ⁢ π 3 = e i ⁢ 2 ⁢ π 3 2 ⁢ e i ⁢ 2 ⁢ π 3
74 68 69 73 3eqtr3i ⊢ 1 = e i ⁢ 2 ⁢ π 3 2 ⁢ e i ⁢ 2 ⁢ π 3
75 divmul3 ⊢ 1 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 2 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 ≠ 0 → 1 e i ⁢ 2 ⁢ π 3 = e i ⁢ 2 ⁢ π 3 2 ↔ 1 = e i ⁢ 2 ⁢ π 3 2 ⁢ e i ⁢ 2 ⁢ π 3
76 75 biimpar ⊢ 1 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 2 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 ∈ ℂ ∧ e i ⁢ 2 ⁢ π 3 ≠ 0 ∧ 1 = e i ⁢ 2 ⁢ π 3 2 ⁢ e i ⁢ 2 ⁢ π 3 → 1 e i ⁢ 2 ⁢ π 3 = e i ⁢ 2 ⁢ π 3 2
77 55 74 76 mp2an ⊢ 1 e i ⁢ 2 ⁢ π 3 = e i ⁢ 2 ⁢ π 3 2
78 1 oveq2i ⊢ 1 O = 1 e i ⁢ 2 ⁢ π 3
79 1 oveq1i ⊢ O 2 = e i ⁢ 2 ⁢ π 3 2
80 77 78 79 3eqtr4ri ⊢ O 2 = 1 O
81 2 oveq1i ⊢ Z 3 = O 1 3 3
82 3nn ⊢ 3 ∈ ℕ
83 cxproot ⊢ O ∈ ℂ ∧ 3 ∈ ℕ → O 1 3 3 = O
84 14 82 83 mp2an ⊢ O 1 3 3 = O
85 81 84 eqtr2i ⊢ O = Z 3
86 85 oveq1i ⊢ O 2 = Z 3 2
87 85 oveq2i ⊢ 1 O = 1 Z 3
88 80 86 87 3eqtr3i ⊢ Z 3 2 = 1 Z 3
89 3t2e6 ⊢ 3 ⋅ 2 = 6
90 89 oveq2i ⊢ Z 3 ⋅ 2 = Z 6
91 2nn0 ⊢ 2 ∈ ℕ 0
92 expmul ⊢ Z ∈ ℂ ∧ 3 ∈ ℕ 0 ∧ 2 ∈ ℕ 0 → Z 3 ⋅ 2 = Z 3 2
93 19 29 91 92 mp3an ⊢ Z 3 ⋅ 2 = Z 3 2
94 90 93 eqtr3i ⊢ Z 6 = Z 3 2
95 3z ⊢ 3 ∈ ℤ
96 exprec ⊢ Z ∈ ℂ ∧ Z ≠ 0 ∧ 3 ∈ ℤ → 1 Z 3 = 1 Z 3
97 19 25 95 96 mp3an ⊢ 1 Z 3 = 1 Z 3
98 88 94 97 3eqtr4i ⊢ Z 6 = 1 Z 3
99 98 oveq2i ⊢ Z 3 + Z 6 = Z 3 + 1 Z 3
100 51 52 99 3eqtr3i ⊢ − 1 = Z 3 + 1 Z 3
101 sqdivid ⊢ Z ∈ ℂ ∧ Z ≠ 0 → Z 2 Z = Z
102 19 25 101 mp2an ⊢ Z 2 Z = Z
103 19 sqcli ⊢ Z 2 ∈ ℂ
104 103 19 25 divreci ⊢ Z 2 Z = Z 2 ⁢ 1 Z
105 102 104 eqtr3i ⊢ Z = Z 2 ⁢ 1 Z
106 15 5 negsubi ⊢ 1 + -2 = 1 − 2
107 5 15 negsubdi2i ⊢ − 2 − 1 = 1 − 2
108 2m1e1 ⊢ 2 − 1 = 1
109 108 negeqi ⊢ − 2 − 1 = − 1
110 106 107 109 3eqtr2i ⊢ 1 + -2 = − 1
111 110 oveq2i ⊢ Z 1 + -2 = Z − 1
112 1z ⊢ 1 ∈ ℤ
113 91 nn0negzi ⊢ − 2 ∈ ℤ
114 expaddz ⊢ Z ∈ ℂ ∧ Z ≠ 0 ∧ 1 ∈ ℤ ∧ − 2 ∈ ℤ → Z 1 + -2 = Z 1 ⁢ Z − 2
115 19 25 112 113 114 mp4an ⊢ Z 1 + -2 = Z 1 ⁢ Z − 2
116 expn1 ⊢ Z ∈ ℂ → Z − 1 = 1 Z
117 19 116 ax-mp ⊢ Z − 1 = 1 Z
118 111 115 117 3eqtr3i ⊢ Z 1 ⁢ Z − 2 = 1 Z
119 exp1 ⊢ Z ∈ ℂ → Z 1 = Z
120 19 119 ax-mp ⊢ Z 1 = Z
121 expnegz ⊢ Z ∈ ℂ ∧ Z ≠ 0 ∧ 2 ∈ ℤ → Z − 2 = 1 Z 2
122 19 25 70 121 mp3an ⊢ Z − 2 = 1 Z 2
123 19 25 sqrecii ⊢ 1 Z 2 = 1 Z 2
124 122 123 eqtr4i ⊢ Z − 2 = 1 Z 2
125 120 124 oveq12i ⊢ Z 1 ⁢ Z − 2 = Z ⁢ 1 Z 2
126 118 125 eqtr3i ⊢ 1 Z = Z ⁢ 1 Z 2
127 105 126 oveq12i ⊢ Z + 1 Z = Z 2 ⁢ 1 Z + Z ⁢ 1 Z 2
128 3 127 eqtri ⊢ A = Z 2 ⁢ 1 Z + Z ⁢ 1 Z 2
129 128 oveq2i ⊢ 3 ⁢ A = 3 ⁢ Z 2 ⁢ 1 Z + Z ⁢ 1 Z 2
130 103 26 mulcli ⊢ Z 2 ⁢ 1 Z ∈ ℂ
131 26 sqcli ⊢ 1 Z 2 ∈ ℂ
132 19 131 mulcli ⊢ Z ⁢ 1 Z 2 ∈ ℂ
133 9 130 132 adddii ⊢ 3 ⁢ Z 2 ⁢ 1 Z + Z ⁢ 1 Z 2 = 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2
134 129 133 eqtri ⊢ 3 ⁢ A = 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2
135 100 134 oveq12i ⊢ - 1 + 3 ⁢ A = Z 3 + 1 Z 3 + 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2
136 15 negcli ⊢ − 1 ∈ ℂ
137 9 28 mulcli ⊢ 3 ⁢ A ∈ ℂ
138 136 137 addcomi ⊢ - 1 + 3 ⁢ A = 3 ⁢ A + -1
139 expcl ⊢ 1 Z ∈ ℂ ∧ 3 ∈ ℕ 0 → 1 Z 3 ∈ ℂ
140 26 29 139 mp2an ⊢ 1 Z 3 ∈ ℂ
141 9 130 mulcli ⊢ 3 ⁢ Z 2 ⁢ 1 Z ∈ ℂ
142 9 132 mulcli ⊢ 3 ⁢ Z ⁢ 1 Z 2 ∈ ℂ
143 50 140 141 142 add42i ⊢ Z 3 + 1 Z 3 + 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2 = Z 3 + 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2 + 1 Z 3
144 135 138 143 3eqtr3i ⊢ 3 ⁢ A + -1 = Z 3 + 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2 + 1 Z 3
145 39 45 144 3eqtri ⊢ − -3 ⁢ A + 1 = Z 3 + 3 ⁢ Z 2 ⁢ 1 Z + 3 ⁢ Z ⁢ 1 Z 2 + 1 Z 3
146 37 38 145 3eqtr4i ⊢ A 3 = − -3 ⁢ A + 1
147 addeq0 ⊢ A 3 ∈ ℂ ∧ -3 ⁢ A + 1 ∈ ℂ → A 3 + -3 ⁢ A + 1 = 0 ↔ A 3 = − -3 ⁢ A + 1
148 147 biimpar ⊢ A 3 ∈ ℂ ∧ -3 ⁢ A + 1 ∈ ℂ ∧ A 3 = − -3 ⁢ A + 1 → A 3 + -3 ⁢ A + 1 = 0
149 35 146 148 mp2an ⊢ A 3 + -3 ⁢ A + 1 = 0