Metamath Proof Explorer


Theorem lognegb

Description: If a number has imaginary part equal to _pi , then it is on the negative real axis and vice-versa. (Contributed by Mario Carneiro, 23-Sep-2014)

Ref Expression
Assertion lognegb A A 0 A + log A = π

Proof

Step Hyp Ref Expression
1 logneg A + log A = log A + i π
2 1 fveq2d A + log A = log A + i π
3 relogcl A + log A
4 pire π
5 crim log A π log A + i π = π
6 3 4 5 sylancl A + log A + i π = π
7 2 6 eqtrd A + log A = π
8 negneg A A = A
9 8 adantr A A 0 A = A
10 9 fveq2d A A 0 log A = log A
11 10 fveqeq2d A A 0 log A = π log A = π
12 7 11 syl5ib A A 0 A + log A = π
13 logcl A A 0 log A
14 13 replimd A A 0 log A = log A + i log A
15 14 fveq2d A A 0 e log A = e log A + i log A
16 eflog A A 0 e log A = A
17 13 recld A A 0 log A
18 17 recnd A A 0 log A
19 ax-icn i
20 13 imcld A A 0 log A
21 20 recnd A A 0 log A
22 mulcl i log A i log A
23 19 21 22 sylancr A A 0 i log A
24 efadd log A i log A e log A + i log A = e log A e i log A
25 18 23 24 syl2anc A A 0 e log A + i log A = e log A e i log A
26 15 16 25 3eqtr3d A A 0 A = e log A e i log A
27 oveq2 log A = π i log A = i π
28 27 fveq2d log A = π e i log A = e i π
29 efipi e i π = 1
30 28 29 eqtrdi log A = π e i log A = 1
31 30 oveq2d log A = π e log A e i log A = e log A -1
32 31 eqeq2d log A = π A = e log A e i log A A = e log A -1
33 26 32 syl5ibcom A A 0 log A = π A = e log A -1
34 17 rpefcld A A 0 e log A +
35 34 rpcnd A A 0 e log A
36 neg1cn 1
37 mulcom e log A 1 e log A -1 = -1 e log A
38 35 36 37 sylancl A A 0 e log A -1 = -1 e log A
39 35 mulm1d A A 0 -1 e log A = e log A
40 38 39 eqtrd A A 0 e log A -1 = e log A
41 40 negeqd A A 0 e log A -1 = e log A
42 35 negnegd A A 0 e log A = e log A
43 41 42 eqtrd A A 0 e log A -1 = e log A
44 43 34 eqeltrd A A 0 e log A -1 +
45 negeq A = e log A -1 A = e log A -1
46 45 eleq1d A = e log A -1 A + e log A -1 +
47 44 46 syl5ibrcom A A 0 A = e log A -1 A +
48 33 47 syld A A 0 log A = π A +
49 12 48 impbid A A 0 A + log A = π