Metamath Proof Explorer


Theorem crreczi

Description: Reciprocal of a complex number in terms of real and imaginary components. Remark in Apostol p. 361. (Contributed by NM, 29-Apr-2005) (Proof shortened by Jeff Hankins, 16-Dec-2013)

Ref Expression
Hypotheses crrecz.1 ⊢ A ∈ ℝ
crrecz.2 ⊢ B ∈ ℝ
Assertion crreczi ⊢ A ≠ 0 ∨ B ≠ 0 → 1 A + i ⁢ B = A − i ⁢ B A 2 + B 2

Proof

Step Hyp Ref Expression
1 crrecz.1 ⊢ A ∈ ℝ
2 crrecz.2 ⊢ B ∈ ℝ
3 1 recni ⊢ A ∈ ℂ
4 3 sqcli ⊢ A 2 ∈ ℂ
5 ax-icn ⊢ i ∈ ℂ
6 2 recni ⊢ B ∈ ℂ
7 5 6 mulcli ⊢ i ⁢ B ∈ ℂ
8 7 sqcli ⊢ i ⁢ B 2 ∈ ℂ
9 4 8 negsubi ⊢ A 2 + − i ⁢ B 2 = A 2 − i ⁢ B 2
10 5 6 sqmuli ⊢ i ⁢ B 2 = i 2 ⁢ B 2
11 i2 ⊢ i 2 = − 1
12 11 oveq1i ⊢ i 2 ⁢ B 2 = -1 ⁢ B 2
13 ax-1cn ⊢ 1 ∈ ℂ
14 6 sqcli ⊢ B 2 ∈ ℂ
15 13 14 mulneg1i ⊢ -1 ⁢ B 2 = − 1 ⁢ B 2
16 10 12 15 3eqtri ⊢ i ⁢ B 2 = − 1 ⁢ B 2
17 16 negeqi ⊢ − i ⁢ B 2 = − − 1 ⁢ B 2
18 13 14 mulcli ⊢ 1 ⁢ B 2 ∈ ℂ
19 18 negnegi ⊢ − − 1 ⁢ B 2 = 1 ⁢ B 2
20 14 mullidi ⊢ 1 ⁢ B 2 = B 2
21 17 19 20 3eqtri ⊢ − i ⁢ B 2 = B 2
22 21 oveq2i ⊢ A 2 + − i ⁢ B 2 = A 2 + B 2
23 3 7 subsqi ⊢ A 2 − i ⁢ B 2 = A + i ⁢ B ⁢ A − i ⁢ B
24 9 22 23 3eqtr3ri ⊢ A + i ⁢ B ⁢ A − i ⁢ B = A 2 + B 2
25 24 oveq1i ⊢ A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2 = A 2 + B 2 A 2 + B 2
26 neorian ⊢ A ≠ 0 ∨ B ≠ 0 ↔ ¬ A = 0 ∧ B = 0
27 sumsqeq0 ⊢ A ∈ ℝ ∧ B ∈ ℝ → A = 0 ∧ B = 0 ↔ A 2 + B 2 = 0
28 1 2 27 mp2an ⊢ A = 0 ∧ B = 0 ↔ A 2 + B 2 = 0
29 28 necon3bbii ⊢ ¬ A = 0 ∧ B = 0 ↔ A 2 + B 2 ≠ 0
30 26 29 bitri ⊢ A ≠ 0 ∨ B ≠ 0 ↔ A 2 + B 2 ≠ 0
31 3 7 addcli ⊢ A + i ⁢ B ∈ ℂ
32 3 7 subcli ⊢ A − i ⁢ B ∈ ℂ
33 4 14 addcli ⊢ A 2 + B 2 ∈ ℂ
34 31 32 33 divasszi ⊢ A 2 + B 2 ≠ 0 → A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2 = A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2
35 30 34 sylbi ⊢ A ≠ 0 ∨ B ≠ 0 → A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2 = A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2
36 divid ⊢ A 2 + B 2 ∈ ℂ ∧ A 2 + B 2 ≠ 0 → A 2 + B 2 A 2 + B 2 = 1
37 33 36 mpan ⊢ A 2 + B 2 ≠ 0 → A 2 + B 2 A 2 + B 2 = 1
38 30 37 sylbi ⊢ A ≠ 0 ∨ B ≠ 0 → A 2 + B 2 A 2 + B 2 = 1
39 25 35 38 3eqtr3a ⊢ A ≠ 0 ∨ B ≠ 0 → A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2 = 1
40 32 33 divclzi ⊢ A 2 + B 2 ≠ 0 → A − i ⁢ B A 2 + B 2 ∈ ℂ
41 30 40 sylbi ⊢ A ≠ 0 ∨ B ≠ 0 → A − i ⁢ B A 2 + B 2 ∈ ℂ
42 31 a1i ⊢ A ≠ 0 ∨ B ≠ 0 → A + i ⁢ B ∈ ℂ
43 crne0 ⊢ A ∈ ℝ ∧ B ∈ ℝ → A ≠ 0 ∨ B ≠ 0 ↔ A + i ⁢ B ≠ 0
44 1 2 43 mp2an ⊢ A ≠ 0 ∨ B ≠ 0 ↔ A + i ⁢ B ≠ 0
45 44 biimpi ⊢ A ≠ 0 ∨ B ≠ 0 → A + i ⁢ B ≠ 0
46 divmul ⊢ 1 ∈ ℂ ∧ A − i ⁢ B A 2 + B 2 ∈ ℂ ∧ A + i ⁢ B ∈ ℂ ∧ A + i ⁢ B ≠ 0 → 1 A + i ⁢ B = A − i ⁢ B A 2 + B 2 ↔ A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2 = 1
47 13 46 mp3an1 ⊢ A − i ⁢ B A 2 + B 2 ∈ ℂ ∧ A + i ⁢ B ∈ ℂ ∧ A + i ⁢ B ≠ 0 → 1 A + i ⁢ B = A − i ⁢ B A 2 + B 2 ↔ A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2 = 1
48 41 42 45 47 syl12anc ⊢ A ≠ 0 ∨ B ≠ 0 → 1 A + i ⁢ B = A − i ⁢ B A 2 + B 2 ↔ A + i ⁢ B ⁢ A − i ⁢ B A 2 + B 2 = 1
49 39 48 mpbird ⊢ A ≠ 0 ∨ B ≠ 0 → 1 A + i ⁢ B = A − i ⁢ B A 2 + B 2