How can I introduce complex numbers to precalculus students?
Edan Maor provides an answer on this post, which gives a nice explanation:
A way to solve polynomials
We came up with equations like $x - 5 = 0$, what is $x$?, and the naturals solved them (easily). Then we asked, "wait, what about $x + 5 = 0$?" So we invented negative numbers. Then we asked "wait, what about $2x = 1$?" So we invented rational numbers. Then we asked "wait, what about $x^2 = 2$?" so we invented irrational numbers.
Finally, we asked, "wait, what about $x^2 = -1$?" This is the only question that was left, so we decided to invent the complex numbers, in particular "imaginary" numbers, to solve it. All the other numbers, at some point, didn't exist and didn't seem "real", but now they're fine. Now that we have complex numbers, we can solve every polynomial, so it makes sense that that's the last place to stop.
Pairs of numbers
This explanation goes the route of redefinition. Tell the listener to forget everything he knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good "realistic" use of pairs of numbers (many exist).
Then, show that in this system, $(0,1) * (0,1) = (-1,0)$, in other words, we've defined a new system, under which it makes sense to say that $\sqrt{-1} = i$, when $i=(0,1)$. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there is an answer to $\sqrt{-1}$.
- Along these lines, see André Nicolas's answer to this post.
The historical explanation
Explain the history of the imaginary numbers. Showing that mathematicians also fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end.
I'm a little rusty, but I think there were certain equations that kept having parts of them which used $\sqrt{-1}$, and the mathematicians kept throwing out the equations since there is no such thing.
Then, one mathematician decided to just "roll with it", and kept working, and found out that all those square roots cancelled each other out.
Amazingly, the answer that was left was the correct answer (he was working on finding roots of polynomials, I think). Which lead him to think that there was a valid reason to use $\sqrt{-1}$, even if it took a long time to understand it.
Another idea comes from Byron Schmuland answer on this Math Overflow thread:
Euclidean Geometry
Use complex numbers to explain Ptolemy's Theorem. For a cyclic quadrilateral with vertices $A,B,C,D$ we have $$|AC|\cdot|BD| = |AB|\cdot|CD|+|BC|\cdot|AD|$$
- Thanks to André Nicolas, whom I'll quote: that mathematician "was Bombelli, showing that for cubics with three distinct real roots, the Cardano Formula involved square roots of negative numbers." (See André Nicolas's comment below.)
I really liked the explanation given by the site BetterExplained. Basically, you think of multiplication of real numbers as a geometric transformation.
Take any real number, and represent it as an arrow on the real line, starting from $0$. If you multiply it by a positive number, you change the arrow's length. If you multiply it by $-1$, you make the arrow point the other way. And multiplication by negative numbers other than $-1$ is just a reflection together with a scaling.
Thinking about multiplication this way, if you start with $1$ and multiply twice by $-1$, you reflect twice and therefore you get back to where you started. This makes sense, since $(-1)^2 = 1$. Now, is there an operation that, applied twice to $1$, gives $-1$? It's clear that no real number will do. But what if you escape the real line for a bit and rotate $1$ twice by $90^\circ$? You get $-1$. So now it turns out that the reflections that negative numbers were responsible for were just rotations by $180^\circ$, but we couldn't tell because we were restricted to just a line.
And now we could say that this rotation by $90^\circ$ deserves a name, so let's call it $i$. Of course, a rotation by $90^\circ$ in the other direction will also do, so we just pick one of them and call it $i$, and the other will be $-i$, because they're mirror images of each other.
The last step is realizing that now we have the whole plane to work with. We can use what we know about vectors and say that any arrow starting at $0$ and ending somewhere in the plane can be represented as a linear combination of $1$ and $i$.
Thinking this way, De Moivre's formula becomes a definition, which makes it clear what complex number multiplication really is (absolute values are multiplied while angles are added), because that's how we started! We wanted $i$ to represent a rotation by $90^\circ$, not some abstract solution of $x^2 = -1$.
One of my favorite introductions to the complex numbers is to not define them with respect to the square root of $-1$! In fact, introduce the complex numbers and the operations therein as just an extension of those of the reals that they all know and love. Then, for instance, define addition and how this works in the plane (as vectors). Then, define multiplication, albeit in the necessarily "strange" way, saying that we need to be able to multiply these 2-tuples. Then proceed to blow their mind that actually $(0,1)$ is the square root of $(-1,0)$! It seems that the typical introduction to the complex number system merely by talking about the square root of $-1$ makes students think that complex numbers don't "exist" or are not important because it addresses a seemingly trivial problem.