Main derivation
S01 · definition
Before
z=2+3i,\ w=4-5i
After
z\leftrightarrow(2,3),\ w\leftrightarrow(4,-5)
Conditions: none
Rule: mko_complex_def · Uses: G1, G2
Leaf basis: mko_complex_def + complex_pair_eq
By the MKO, a complex number a+bi is determined by its real part a and imaginary coefficient b and corresponds to the ordered pair (a,b). z has real part 2 and imaginary coefficient 3; w=4−5i=4+(−5)i has imaginary coefficient −5: the minus sign belongs to the coefficient and must not be dropped.
S02 · equivalent
Before
z+w
After
(2+4)+(3+(-5))i=6-2i\leftrightarrow(6,-2)
Conditions: none
Rule: complex_addsub · Uses: S01, complex_addsub
Leaf basis: complex_addsub + numeric_evaluation
Addition is componentwise: real part 2+4=6, imaginary coefficient 3+(−5)=−2, so z+w=6−2i, the ordered pair (6,−2).
S03 · equivalent
Before
z-w
After
(2-4)+(3-(-5))i=-2+8i\leftrightarrow(-2,8)
Conditions: none
Rule: complex_addsub · Uses: S01, complex_addsub
Leaf basis: complex_addsub + add_inverse + numeric_evaluation
Subtraction is also componentwise: real part 2−4=−2, imaginary coefficient 3−(−5)=3+5=8. Subtracting a negative number adds its opposite; this is where signs are most easily lost.
S04 · equivalent
Before
(a+bi)(c+di)
After
(ac-bd)+(ad+bc)i
Conditions: a,b,c,d\in\mathbb{R}
Rule: L_complex_mul · Uses: mko_complex_def, i_squared
Leaf basis: distributive + mul_assoc + mul_comm + i_squared + add_comm
The MKO says multiplication follows from distributivity and i²=−1; it is not componentwise. First prove the general formula for arbitrary reals a, b, c, d (Lemma L_complex_mul) in sub-steps S04.1 to S04.3; the products below are then computed with it.
S05 · theorem_application
Before
zw=(2+3i)(4+(-5)i)
After
zw=(2\cdot 4-3\cdot(-5))+(2\cdot(-5)+3\cdot 4)i
Conditions: a=2,\ b=3,\ c=4,\ d=-5
Rule: L_complex_mul · Uses: S01, S04
Leaf basis: L_complex_mul + forall_elim
Substitute a=2, b=3, c=4, d=−5 into the lemma. d must be −5, not 5: the imaginary coefficient of w is substituted together with its sign.
S06 · equivalent
Before
(2\cdot 4-3\cdot(-5))+(2\cdot(-5)+3\cdot 4)i
After
23+2i
Conditions: none
Rule: numeric_evaluation · Uses: S05
Leaf basis: numeric_evaluation
Real part: 8−(−15)=8+15=23; imaginary coefficient: −10+12=2. So zw=23+2i.
S07 · equivalent
Before
(2+3i)(4-5i)
After
8-10i+12i-15i^2=8+2i+15=23+2i
Conditions: none
Rule: i_squared · Uses: S01, distributive, i_squared
Leaf basis: distributive + mul_assoc + mul_comm + i_squared + numeric_evaluation
Cross-check by direct expansion without the formula: distributivity gives 2·4+2·(−5i)+3i·4+3i·(−5i)=8−10i+12i−15i²; i²=−1 turns −15i² into +15; collecting gives 23+2i, the same as S06.
S08 · equivalent
Before
zw=23+2i
After
zw\leftrightarrow(23,2)
Conditions: none
Rule: complex_pair_eq · Uses: S06, complex_pair_eq
Leaf basis: complex_pair_eq
By G2 the answer is written as the ordered pair (real part, imaginary coefficient)=(23,2). The order is semantic data: (2,23) stands for 2+23i, a different complex number.
S09 · implies
Before
(23,2)\ \text{vs}\ (2,23)After
(23,2)\neq(2,23)
Conditions: none
Rule: complex_pair_eq · Uses: S08, complex_pair_eq
Leaf basis: complex_pair_eq + numeric_evaluation
Two ordered pairs are equal only if the real parts agree and the imaginary coefficients agree; 23≠2, so (23,2)≠(2,23). Swapping the real and imaginary parts gives a different complex number, which is the common mistake the MKO names.
S10 · definition
Before
23\in\mathbb{R}After
23=23+0i\leftrightarrow(23,0)
Conditions: none
Rule: real_embedding · Uses: real_embedding
Leaf basis: real_embedding
To decide whether zw equals the real number 23, first embed 23 in the complex numbers: 23=23+0i, the ordered pair (23,0). The comparison must take place in one set, ℂ.
S11 · equivalent
Before
zw=23
After
23=23\ \land\ 2=0
Conditions: none
Rule: complex_pair_eq · Uses: S08, S10, complex_pair_eq
Leaf basis: complex_pair_eq + and_intro_elim
zw=23 is equivalent to (23,2)=(23,0), which in turn is equivalent to both components agreeing: 23=23 and 2=0.
S12 · implies
Before
23=23\ \land\ 2=0
After
zw\neq 23
Conditions: none
Rule: contradiction · Uses: S11
Leaf basis: contradiction + and_intro_elim + numeric_evaluation
2=0 is false, so the conjunction is false, and by the equivalence S11, zw=23 fails. This is the MKO's third paragraph: a complex number with nonzero imaginary coefficient equals no real number, whatever its real part.
S13 · theorem_application
Before
wz=(4+(-5)i)(2+3i)
After
wz=(4\cdot 2-(-5)\cdot 3)+(4\cdot 3+(-5)\cdot 2)i=23+2i
Conditions: a=4,\ b=-5,\ c=2,\ d=3
Rule: L_complex_mul · Uses: S04, G1
Leaf basis: L_complex_mul + numeric_evaluation
Compute wz in the other order as a further check: the lemma gives real part 8+15=23 and imaginary coefficient 12−10=2, the same as zw, consistent with commutativity of multiplication in ℂ.
S14 · implies
Before
z=2+3i,\ w=4-5i
After
z+w\leftrightarrow(6,-2),\ z-w\leftrightarrow(-2,8),\ zw\leftrightarrow(23,2),\ zw\neq 23
Conditions: none
Rule: and_intro_elim · Uses: S02, S03, S08, S12
Leaf basis: and_intro_elim
Combine the four conclusions: sum (6,−2), difference (−2,8), product (23,2), and the product is not the real number 23.
Error controls and negative controls
Forgetting i²=−1
One mistake the MKO names is forgetting i²=−1 in multiplication. Treating −15i² as −15 (that is, taking i² as 1) gives (8−15)+2i=−7+2i; dropping the i² term gives 8+2i. Correctly, −15i²=−15·(−1)=+15 and the real part is 23. S04.3 shows that the −bd in the general formula is exactly this sign.
Swapping the real and imaginary parts
Another mistake identified in the MKO is swapping the real and imaginary parts. The answer (23,2) means 23+2i; writing (2,23) gives 2+23i, a different number (S09). The order of the pair is semantic data and is not guessed from positions in a picture.
Multiplication is not componentwise
Addition and subtraction are componentwise, but multiplication is not: (2,3)·(4,−5) is not (2·4, 3·(−5))=(8,−15). In the correct formula the real part ac−bd mixes both real parts and both imaginary coefficients, and the imaginary coefficient ad+bc is the sum of the cross terms.
- (2+3i)(4−5i)=8−15i, because multiplication is componentwise →
reject — Multiplication follows from distributivity and i²=−1; the correct product is 23+2i. - (2+3i)(4−5i)=−7+2i →
reject — This takes i² as +1; in fact −15i²=+15 and the real part is 8+15=23. - The ordered pair of zw is (2,23) →
reject — The order is (real part, imaginary coefficient); zw=23+2i corresponds to (23,2), and (2,23)≠(23,2). - 23+2i has real part 23, so it equals the real number 23 →
reject — 23 embeds as (23,0), and equality would also need 2=0, which fails.