Main derivation
S01 · definition
Before
z=a+bi\leftrightarrow(a,b)
After
\overline{z}=a-bi\leftrightarrow(a,-b),\ |z|^2=a^2+b^2Conditions: none
Rule: mko_modconj_def · Uses: G1, mko_modconj_def
Leaf basis: mko_modconj_def + def_intro
By the MKO, conjugation changes the sign of the imaginary coefficient only and leaves the real part alone: (a,b) becomes (a,−b); the modulus squared is the sum of squares of the components, a²+b². The modulus and its square are different notions: the modulus is the nonnegative square root of the modulus squared.
S02 · equivalent
Before
z\overline{z}=(a,b)\cdot(a,-b)After
(a^2+b^2,\,0)
Conditions: none
Rule: complex_mul_formula · Uses: S01, complex_mul_formula
Leaf basis: complex_mul_formula + complex_pair_eq
Compute z·z̄ by the multiplication formula with c=a and d=−b: the real part is S02.1 and the imaginary coefficient is S02.2.
S03 · implies
Before
z\overline{z}=(a^2+b^2,\,0)After
z\overline{z}=|z|^2\in\mathbb{R},\ |z|^2\ge 0Conditions: none
Rule: real_embedding · Uses: S02, real_embedding, square_nonneg
Leaf basis: real_embedding + square_nonneg + order_basics
(a²+b²,0) is exactly the embedding of the real number a²+b²=|z|², so z·z̄ equals the real embedding of the modulus squared, as the MKO says. Also a²≥0 and b²≥0, so |z|²≥0, with equality only when a=b=0.
S04 · equivalent
Before
\overline{z+w}=\overline{(a+c)+(b+d)i}After
(a+c)-(b+d)i=(a-bi)+(c-di)=\overline{z}+\overline{w}Conditions: none
Rule: complex_addsub · Uses: G1, complex_addsub, S01
Leaf basis: complex_addsub + complex_pair_eq + distributive
z+w has imaginary coefficient b+d, which conjugation turns into −(b+d)=−b−d; adding z̄+w̄ componentwise gives imaginary coefficient −b+(−d) as well. Both real parts are a+c, so conjugation preserves addition.
S05 · equivalent
Before
\overline{zw}After
\overline{z}\cdot\overline{w}Conditions: none
Rule: L_conj_mul · Uses: S01, complex_mul_formula
Leaf basis: complex_mul_formula + complex_pair_eq + neg_mul
Conjugation preserves multiplication (Lemma L_conj_mul). The proof is in S05.1 and S05.2: compute the ordered pair of each side and compare components.
S06 · equivalent
Before
|zw|^2
After
|z|^2\,|w|^2
Conditions: none
Rule: L_modsq_mul · Uses: S03, S05
Leaf basis: real_embedding + mul_assoc + mul_comm + complex_pair_eq
From S03 and S05, the modulus squared is multiplicative (Lemma L_modsq_mul), proved in S06.1 and S06.2. N5 cross-checks independently with the polynomial identity (ac−bd)²+(ad+bc)²=(a²+b²)(c²+d²).
S07 · equivalent
Before
z=3-4i\leftrightarrow(3,-4)
After
\overline{z}=3+4i\leftrightarrow(3,4)Conditions: none
Rule: mko_modconj_def · Uses: G2, S01
Leaf basis: mko_modconj_def
Only the imaginary coefficient −4 changes sign to 4; the real part 3 is unchanged. Writing −3−4i or −3+4i is wrong (the first changes the real part, the second is −z).
S08 · implies
Before
|z|^2=3^2+(-4)^2,\ z\overline{z}=(3,-4)(3,4)After
|z|^2=25,\ z\overline{z}=(3\cdot3-(-4)\cdot4,\,3\cdot4+(-4)\cdot3)=(25,0)Conditions: none
Rule: complex_mul_formula · Uses: S07, S02, complex_mul_formula
Leaf basis: complex_mul_formula + numeric_evaluation
The modulus squared is 9+16=25. Computing z·z̄ directly by the formula: real part 9+16=25, imaginary coefficient 12−12=0, giving (25,0), the real embedding of 25, in line with the general results S02 and S03.
S09 · implies
Before
|z|^2=25
After
|z|=\sqrt{25}=5\neq 25Conditions: 5\ge 0
Rule: sqrt_def · Uses: S08, modulus_def, sqrt_def
Leaf basis: sqrt_def + modulus_def + numeric_evaluation
The modulus is the nonnegative square root of the modulus squared. Since 5≥0 and 5²=25, uniqueness of the square root gives |z|=5. The modulus squared 25 and the modulus 5 are different numbers, which is the mistake the MKO names.
S10 · implies
Before
w=1+2i
After
\overline{w}=1-2i,\ |w|^2=5,\ |w|=\sqrt5\notin\mathbb{Z}Conditions: none
Rule: mko_modconj_def · Uses: G2, G3, S01
Leaf basis: mko_modconj_def + order_basics + case_analysis + numeric_evaluation
w̄=(1,−2) and |w|²=1+4=5. 5 is not a perfect square: a nonnegative integer n≤2 has n²≤4 and n≥3 has n²≥9, so no integer squares to 5. Hence |w|=√5 is not an integer; by G3 the exact modulus squared 5 is kept and no decimal such as 2.236 is written.
S11 · equivalent
Before
zw=(3,-4)(1,2)
After
(3\cdot1-(-4)\cdot2,\,3\cdot2+(-4)\cdot1)=(11,2)
Conditions: none
Rule: complex_mul_formula · Uses: G2, complex_mul_formula
Leaf basis: complex_mul_formula + numeric_evaluation
Real part 3+8=11, imaginary coefficient 6−4=2, so zw=11+2i.
S12 · implies
Before
\overline{zw}\ \text{vs}\ \overline{z}\,\overline{w}=(3,4)(1,-2)After
(11,-2)=(3\cdot1-4\cdot(-2),\,3\cdot(-2)+4\cdot1)
Conditions: none
Rule: complex_mul_formula · Uses: S11, S07, S10, S05
Leaf basis: complex_mul_formula + complex_pair_eq + numeric_evaluation
Left: (zw)‾=(11,−2). Right: z̄·w̄=(3,4)(1,−2), real part 3+8=11, imaginary coefficient −6+4=−2. The two sides agree, an instance of S05.
S13 · implies
Before
|zw|^2=11^2+2^2
After
125=25\cdot5=|z|^2|w|^2
Conditions: none
Rule: L_modsq_mul · Uses: S11, S08, S10, S06
Leaf basis: L_modsq_mul + numeric_evaluation
121+4=125 and |z|²|w|²=25·5=125 agree, an instance of S06. Even though |w| is not an integer, the modulus squares stay exact integers.
S14 · implies
Before
\overline{z+w}=\overline{(4,-2)}After
(4,2)=(3,4)+(1,-2)=\overline{z}+\overline{w}Conditions: none
Rule: complex_addsub · Uses: G2, S04, S07, S10
Leaf basis: complex_addsub + numeric_evaluation
z+w=(3+1, −4+2)=(4,−2), whose conjugate is (4,2); z̄+w̄=(3+1, 4−2)=(4,2). The sides agree, an instance of S04.
S15 · implies
Before
z=a+bi,\ w=c+di
After
z\overline{z}=(|z|^2,0),\ \overline{z+w}=\overline{z}+\overline{w},\ \overline{zw}=\overline{z}\,\overline{w},\ |zw|^2=|z|^2|w|^2Conditions: none
Rule: and_intro_elim · Uses: S03, S04, S05, S06
Leaf basis: and_intro_elim + forall_intro
Combine the four general results; they hold for all real components, and the integer examples of S07 to S14 are only checks.
Substeps and lemmas
S02.1 · equivalent
Before
\operatorname{Re}(z\overline{z})=a\cdot a-b\cdot(-b)After
a^2+b^2
Conditions: none
Rule: neg_mul · Uses: complex_mul_formula, neg_mul
Leaf basis: neg_mul + add_inverse + power_laws
In the real part ac−bd put c=a and d=−b: a·a−b·(−b). Since b·(−b)=−b², subtracting −b² adds b², giving a²+b². The plus sign comes from the product of i²=−1 with the conjugate's minus sign.
S02.2 · equivalent
Before
\operatorname{Im}(z\overline{z})=a\cdot(-b)+b\cdot aAfter
0
Conditions: none
Rule: mul_comm · Uses: complex_mul_formula, mul_comm
Leaf basis: mul_comm + neg_mul + add_inverse
In the imaginary coefficient ad+bc put d=−b and c=a: a·(−b)+b·a=−ab+ab=0 (commutativity gives ba=ab). So z·z̄ has zero imaginary part and is real.
S05.1 · equivalent
Before
\overline{zw}=\overline{(ac-bd)+(ad+bc)i}After
(ac-bd,\,-(ad+bc))
Conditions: none
Rule: complex_mul_formula · Uses: complex_mul_formula, S01
Leaf basis: complex_mul_formula + mko_modconj_def
First write zw=(ac−bd)+(ad+bc)i by the multiplication formula, then conjugate: the real part stays and the imaginary coefficient changes sign.
S05.2 · equivalent
Before
\overline{z}\cdot\overline{w}=(a,-b)\cdot(c,-d)After
(ac-(-b)(-d),\,a(-d)+(-b)c)=(ac-bd,\,-(ad+bc))
Conditions: none
Rule: neg_mul · Uses: S05.1, complex_mul_formula, neg_mul
Leaf basis: neg_mul + distributive + complex_pair_eq
Apply the multiplication formula to (a,−b) and (c,−d): real part ac−(−b)(−d)=ac−bd (two minus signs give a plus); imaginary coefficient a(−d)+(−b)c=−ad−bc=−(ad+bc). Both components match S05.1, so by equality of ordered pairs (zw)‾=z̄·w̄. S05.1 and S05.2 together prove Lemma L_conj_mul.
S06.1 · equivalent
Before
(|zw|^2,0)=(zw)\,\overline{zw}After
(zw)(\overline{z}\,\overline{w})=(z\overline{z})(w\overline{w})Conditions: none
Rule: mul_assoc · Uses: S03, S05, mul_assoc, mul_comm
Leaf basis: eq_replace + mul_assoc + mul_comm
Apply S03 to the complex number zw: (zw)·(zw)‾ is the embedding of |zw|². Replace (zw)‾ by z̄·w̄ using S05, then regroup the four factors as (z z̄)(w w̄) by associativity and commutativity of multiplication in ℂ.
S06.2 · implies
Before
(z\overline{z})(w\overline{w})=(|z|^2,0)(|w|^2,0)After
(|z|^2|w|^2,\,0)\ \Rightarrow\ |zw|^2=|z|^2|w|^2
Conditions: none
Rule: real_embedding · Uses: S06.1, S03, real_embedding, complex_pair_eq
Leaf basis: real_embedding + complex_pair_eq
By S03, z z̄ and w w̄ are the embeddings of |z|² and |w|²; the product of two embedded reals is (|z|²|w|²,0). Comparing real parts gives |zw|²=|z|²|w|². S06.1 and S06.2 together prove Lemma L_modsq_mul.