1
Treat i as a variable until the last step
Expand (3 + 2i)(4 - 5i) exactly as you would expand (3 + 2t)(4 - 5t). Only then replace i² by -1. Nothing about the manipulation is new; the substitution is the only thing that is.
Lesson · Quadratics and Complex Numbers · Complex numbers
Operations with Complex Numbers
Addition and subtraction collect like terms, multiplication is the distributive law followed by a single substitution, and division is multiplication by the conjugate. An answer is finished only when it is written as a + bi with a and b real.
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi)(c + di) = (ac - bd) + (ad + bc)i because i² = -1
conjugate of a + bi = a - bi
(a + bi)(a - bi) = a² + b² real, and 0 only when a = b = 0
Three operations and the conjugate
Core ideas
What this lesson establishes
Complex arithmetic is ordinary algebra in one letter, with one rule applied at the end and one form the answer has to reach.
2
Standard form is the finishing line
An answer is a + bi with a and b real. 2/(1 + i) is a complex number but is not yet in that form. Two complex numbers are compared part by part — and only for equality: there is no ordering of the complex numbers that respects the arithmetic, so 3 + 2i < 5 is not a false statement, it is not a statement at all.
3
The conjugate turns a denominator real
(a + bi)(a - bi) = a² + b², with no i left. That one fact is the whole of complex division, and it is the same cancellation that cleared 2 + √3 from a denominator in Rational and Radical Expressions.
The material
Arithmetic in a + bi
Four operations, and the one substitution they all come down to.
“Complex Numbers” defined i by i² = -1 and called a + bi a complex number, with real part a and imaginary part b, both of them real numbers. The arithmetic below is arranged so that every law from Foundations of Algebra — commutativity, associativity, distribution — keeps working unchanged. That was the design goal when the system was built, and it is why there is so little to learn here.
Definition · Addition and subtraction
(a + bi) + (c + di) = (a + c) + (b + d)i, and (a + bi) - (c + di) = (a - c) + (b - d)i. Real parts combine with real parts, imaginary parts with imaginary parts.
Nothing is being asserted beyond collecting like terms: bi and di are both multiples of i, so they add the way 3t and 5t do. (2 - 7i) - (5 + 3i) = -3 - 10i, and the minus sign has to reach both parts of the second bracket — the same place a sign goes missing when polynomials are subtracted in Polynomials and Factoring.
Definition · Multiplication
(a + bi)(c + di) = (ac - bd) + (ad + bc)i. Do not memorise that. Expand the two brackets, replace i² by -1, and collect; the formula is what that always produces.
(a + bi)(c + di) = ac + adi + bci + bd·i²
= ac + adi + bci - bd since i² = -1
= (ac - bd) + (ad + bc)i
Example · A product in full
(3 + 2i)(4 - 5i) = 12 - 15i + 8i - 10i². Now -10i² is -10 · (-1) = +10, so the answer is 22 - 7i.
The +10 is where complex multiplication goes wrong. Subtracting 10i² adds 10, and an answer that is 20 out from the right one usually has that step in it.
Powers of i
i⁰ = 1 i¹ = i i² = -1 i³ = -i
i⁴ = 1 i⁵ = i i⁶ = -1 i⁷ = -i
i⁴ = (i²)² = (-1)² = 1, so the four values repeat for ever
So iⁿ depends only on the remainder of n on division by 4. i²³ has 23 = 4·5 + 3, remainder 3, so i²³ = -i. Written out: i²³ = (i⁴)⁵ · i³ = 1 · i³ = -i.
One law from Foundations of Algebra does not extend. √a · √b = √(ab) is stated for non-negative a and b, and it fails here: √-4 · √-9 is 2i · 3i = 6i² = -6, not √36 = 6. Convert every root of a negative number to i first, then multiply.
Definition · Complex conjugate
The conjugate of a + bi is a - bi: same real part, opposite imaginary part. Conjugating twice returns the original number, and a number equals its own conjugate exactly when b = 0, that is, exactly when it is real.
Theorem · The conjugate product is real
For real a and b, (a + bi)(a - bi) = a² + b². This is a real number, it is never negative, and it is 0 only when a = 0 and b = 0.
Expand: (a + bi)(a - bi) = a² - abi + abi - b²i². The two cross terms are negatives of one another and cancel.
The last term is -b²i² = -b²(-1) = b², so what remains is a² + b², with no i in it.
a and b are real, so a² and b² are each at least 0 and the sum is at least 0. A sum of two non-negative numbers is 0 only when both are 0, which forces a = 0 and b = 0.
□
That is what makes division work. To divide by c + di, multiply above and below by c - di. Multiplying a fraction by (c - di)/(c - di) is multiplying it by 1, so the value is untouched — only the form changes, and the new denominator c² + d² is an ordinary real number you can divide by.
Example · A quotient
(4 + i)/(2 + 3i). The conjugate of the denominator is 2 - 3i. On top: (4 + i)(2 - 3i) = 8 - 12i + 2i - 3i² = 11 - 10i. Underneath: (2 + 3i)(2 - 3i) = 4 + 9 = 13. So the quotient is 11/13 - (10/13)i.
Theorem · Every non-zero complex number has a reciprocal
If a + bi ≠ 0 then a² + b² > 0, and 1/(a + bi) = (a - bi)/(a² + b²), which is again of the form p + qi with p and q real. So the complex numbers are closed under all four operations: adding, subtracting, multiplying and dividing them never leads outside the system.
Interactive
Arithmetic with complex numbers
Add, multiply, conjugate, and divide by the conjugate
Method
Dividing one complex number by another
Four steps. The second is the whole method; the rest is arithmetic.
1
Write the quotient as a fraction
(3 + 2i) ÷ (1 - 4i) becomes (3 + 2i)/(1 - 4i). Nothing can be cancelled yet, and cancelling a term rather than a factor is as illegal here as it was in Rational and Radical Expressions.
2
Multiply top and bottom by the conjugate of the denominator
The conjugate of 1 - 4i is 1 + 4i. You are multiplying by (1 + 4i)/(1 + 4i), which is 1, so the value cannot change.
3
Expand both products and replace i² by -1
The denominator becomes 1 + 16 = 17: real, with the cross terms gone. That cancellation is the only reason the conjugate is the right thing to multiply by rather than any other complex number.
4
Split into two real fractions
(p + qi)/r = p/r + (q/r)i. Keep exact fractions — -5/17 is the answer, and a decimal is a rounding of it.
Complex arithmetic
Score 0 / 0Choose the answer you can justify.
(3 + 2i) divided by (1 - 4i)
The denominator is 1 - 4i, so the conjugate in play is 1 + 4i. Everything else is expansion.
3 + 2i 3 + 2i 1 + 4i
------ = ------ · ------ multiplying by 1
1 - 4i 1 - 4i 1 + 4i
top: (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i²
= 3 + 14i - 8
= -5 + 14i
bottom: (1 - 4i)(1 + 4i) = 1 + 4i - 4i - 16i²
= 1 + 16
= 17
-5 + 14i 5 14
-------- = - -- + -- i standard form
17 17 17
check: (-5/17 + (14/17)i)(1 - 4i)
real part: -5/17 + 56/17 = 51/17 = 3 ✓
imaginary: 20/17 + 14/17 = 34/17 = 2 ✓
The bottom lost its i because +4i and -4i cancel. That is the one thing the conjugate is chosen for; multiplying above and below by 1 - 4i instead would give (1 - 4i)² = -15 - 8i, which is no improvement.
The last line is standard form. (-5 + 14i)/17 is the same number, but a real part and an imaginary part cannot be read off it, so it is not yet an answer.
For a faded quotient, put (5 - 2i)/(1 + i) in standard form. The supplied multiplier is (1 - i)/(1 - i). Expand both products, replace i², split the real denominator across both numerator terms and multiply your result by 1 + i as a check. Compare with 3/2 - (7/2)i; an answer of 7/2 - (7/2)i comes from replacing 2i² by +2 instead of -2.
Common mistakes
1
Leaving i² in the answeri² is not a term to be carried, it is the number -1. An expression still containing i² is unfinished, and the error it hides is usually double the size of that term: -10i² is +10, so getting it wrong moves the answer by 20.
2
Multiplying only the denominator by the conjugateThat changes the value of the fraction. The legal move is multiplying by (c - di)/(c - di), which is 1; multiplying the bottom alone is multiplying the fraction by 1/(c - di), which is a different number.
3
Using √a · √b = √(ab) on negatives√-4 · √-9 is 2i · 3i = -6, not 6. The radical law assumes non-negative radicands, and this is the one place in the course where quoting a law outside its hypotheses gives a wrong sign rather than a wrong form.
Check your understanding
✓
Finish when every answer you write is in the form a + bi.
Multiply two complex numbers, divide two complex numbers and reduce a power of i, without looking anything up. The check on a quotient is multiplying it back by the divisor: three lines, and it catches the sign slips this arithmetic invites.
Note: The conjugate is doing here exactly what it did in Rational and Radical Expressions when a denominator like 2 + √3 had to be cleared: multiply by the thing that makes the cross terms cancel. “Complex Roots of Quadratics” turns it into a statement about solutions — a quadratic with real coefficients that has one non-real root has that root's conjugate as its other one.