Multiplying Complex Numbers
Chapter Eight
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 17 to 18 of 303
In one line
Multiply out the brackets as you always have, then replace i squared by minus one, which throws one term from the imaginary part into the real part.
Doing it from first principles
Take two complex numbers and expand the product term by term.
(a + i b)(c + i d) = a c + i a d + i b c + i^2 b d
Now the only new step: i squared is minus one, so the last term is minus bd, and it is real.
(a + i b)(c + i d) = (a c - b d) + i(a d + b c)
That is the general formula, and it is worth understanding rather than memorising. Two things are happening:
- The real part is the product of the real parts minus the product of the imaginary parts. The minus sign is the i squared.
- The imaginary part is the two cross terms added.
The point most students miss
Look at what just happened to the last term. It began life as an imaginary term, i b times i d, and it ended up in the real part with its sign changed.
That is why multiplication of complex numbers is not two separate multiplications the way addition was two separate additions. The parts are genuinely coupled, and it is the coupling that makes complex numbers interesting. It is also exactly the coupling that produces a rotation, as the chapter on multiplication on the Argand diagram shows.
Worked
(2 + 3i)(4 + 5i) = 8 + 10i + 12i + 15i^2 = -7 + 22i
(1 + i)(1 - i) = 1 - i^2 = 2
(3 - 2i)(3 + 2i) = 9 - 4i^2 = 13
(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i
i(3 - 4i) = 3i - 4i^2 = 4 + 3i
Four of those five are patterns worth recognising on sight.
The second and third are a number times its conjugate, and the answer is real both times. That is the whole reason the conjugate exists, and the next chapter is about it.
The fourth is a square, so the usual expansion of (p + q) squared applies with q = i, and the middle term is the one to watch.
The fifth shows multiplication by i doing something very specific: 3 minus 4i became 4 plus 3i. The parts have swapped and one sign has changed. Plot both points and you will see a quarter turn anticlockwise, which is the geometric fact hiding in the cycle of the powers of i.
Squares and cubes
There is nothing new here, only care. Use the ordinary expansions and reduce every power of i at the end.
Multiplying Complex Numbers
(1 + i)^2 = 1 + 2i + i^2 = 2i
(1 + i)^3 = (1 + i)(2i) = 2i + 2i^2 = -2 + 2i
(1 + i)^4 = (2i)^2 = 4i^2 = -4
The second line shows the move that saves time: do not expand a cube from scratch, use the square you already have.
That (1 + i) to the fourth is minus four is a fact worth carrying. It says (1 + i) to the eighth is sixteen, and so on, which makes an otherwise nasty question a single line. The chapter on De Moivre's theorem gives the general way of doing this for any complex number and any power.
The properties, which are set as a theory question
| Property | Statement |
|---|---|
| Commutative | z1 z2 = z2 z1 |
| Associative | (z1 z2) z3 = z1 (z2 z3) |
| Identity | 1 z = z |
| Distributive over addition | z1 (z2 + z3) = z1 z2 + z1 z3 |
Each is proved by expanding both sides with the general formula and comparing the two parts. Commutativity, for instance, is this.
(a + i b)(c + i d) = (a c - b d) + i(a d + b c)
(c + i d)(a + i b) = (c a - d b) + i(c b + d a)
The real parts agree because real multiplication is commutative, and the imaginary parts agree because real addition is as well.
The one property that fails, and why it matters
Real numbers have the property that a product is zero only if one of the factors is zero, and complex numbers keep that. But they lose something else entirely: there is no sensible way to say that one complex number is greater than another, so you cannot argue about products by saying "both factors are positive". Any inequality argument in this paper has to be made about moduli, which are real, and never about the complex numbers themselves.
Check yourself
| Question | Answer |
|---|---|
| (3 + i)(2 - i) | (3 + i)(2 - i) = 7 - i |
| (1 - 2i)(1 + 2i) | (1 - 2i)(1 + 2i) = 5 |
| (2 - 3i)^2 | (2 - 3i)^2 = -5 - 12i |
| i(1 + i) | i(1 + i) = -1 + i |
| (1 + i)^4 | (1 + i)^4 = -4 |
| (1 - i)^4 | (1 - i)^4 = -4 |
The last two coming out the same is not a coincidence: 1 minus i is the conjugate of 1 plus i, and taking a conjugate and then a power gives the conjugate of the power. Since minus four is real, it is its own conjugate.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.