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Today's extra explains how complex numbers were invented. You already know how to solve equations like

3

*x*+ 4 = 10.

There are three numbers in that equation: 3, 4 and 10, and the solution is

*x*= 2. All of those numbers are integers; that is, the kinds of numbers we use to count objects. But if the equation had been

3

*x*+ 4 = 9,

you'd get

*x*=

^{5}/

_{3}. What's that? It's not a number we use to count things. It's a new kind of number - call it a fraction or a decimal (

^{5}/

_{3}= 1.6666...). Notice that the solution of an equation with integer coefficients led to a non-integer solution. The non-integer solution was sort of forced on us. We either had to admit that some numbers aren't integers, or else not be able to solve what appears to be a very simple equation.

If we consider a more complicated equation, a new kind of number is forced on us. Look at

*x*

^{2}+

*x*= -1.

Even if you don't know the formula for solving equations like this, you can see pretty quickly by trial and error that no kind of counting number is going to solve this equation. The answer can't be positive or less than -1, because for numbers like that,

*x*

^{2}+

*x*is positive. What about -

^{1}/

_{2}? Nope, that gives -

^{1}/

_{4}. So fractions aren't working either.

It looks like we need a number whose square is negative. But a positive squared is positive, and a negative squared is...positive.

We start to get out of it by inventing a new number, called

*i*, that has the property

*i*

^{2}= -1.

*i*stands for "imaginary." But if you try

*i*in the equation, you see another problem:

*i*

^{2}+

*i*would be -1 +

*i*, but the right-hand side is plain old 1. We need a way to get rid of the imaginary part. So let's try numbers of the form

*a + bi*

where

*a*and

*b*are real (not imaginary) numbers. Here's where you have to do some work on your own. Show that if we substitute

*x*=

*a + bi*into the equation, we get

*a*

^{2}-

*b*

^{2}+ 2

*abi*+

*a*+

*bi*= -1

(Hint: Use FOIL and the fact that

*i*

^{2}= -1.)

It looks like we have two unknowns,

*a*and

*b*, but only one equation. That is usually trouble, but we have a trick. For two complex numbers to be equal their real parts have to be equal and their imaginary parts have to be equal. It's the only way. So we really have two equations! Breaking the equation into its real and imaginary parts,

*a*

^{2}-

*b*

^{2}+

*a*= -1

2

*ab*+

*b*= 0

Notice that the second equation has zero on the right, because the original equation above didn't have an imaginary part on the right. Said differently, the right-hand side of the original equation is -1+0

*i*.

From the second one we get

*a*= -

^{1}/

_{2}. We're halfway done! If we then substitute that into the first equation, we get

^{1}/

_{4}-

*b*

^{2}-

^{1}/

_{2}= -1

which you can show has the solution

*b*= √(

^{3}/

_{4}).

And, don't forget about the other solution (there are two square roots to any number except zero):

*b*= -√(

^{3}/

_{4}).

So, we've got two solutions to our equation:

*x*= -1/2+

*i*√(

^{3}/

_{4}) and

*x*= -1/2-

*i*√(

^{3}/

_{4}).

We had to invent a whole new kind of number, a

*complex number*, to solve this equation whose coefficients are plain old integers. A number that is a

*real number*plus an

*imaginary number*is called a

*complex number*. We weren't very careful about defining what a real number is, but I think you have the idea by now. It's a "not imaginary" number. Real numbers include integers, fractions and those non-repeating decimals we call irrational numbers.

Your challenge: Solve the equation

*x*

^{2}-

*x*= -1

Now that I'm done writing this, I don't know how many more I'm going to do this way. Ordinarily I would write math using LaTeX. In HTML, it is a real chore and it looks terrible. Maybe I can find a way to attach a PDF to a blog post. There are also some LaTeX to HTML converters out there, but they never work very well.

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