It's finally come to this, has it? We've run out of actual numbers to throw at you, so now we're just going to make some numbers up? Not really. Imaginary numbers, despite the name, are totally legitimate numbers in their own right. It's just that all of the numbers we've worked with so far are the real numbers, and what else are you going to call something that isn't "real?" Calling them fake numbers wouldn't really help things.

Okay, so what are imaginary numbers? We'll answer that with another question: what is ? It should be a number that, when squared, equals -1, but any positive or negative real number will be positive when squared. We define a new type of number, called i, and state that . You can see that i^{2} = -1, just like we wanted.

Sample Problem

What are the first 8 powers of i equal to? That is, i^{1}, i^{2}, i^{3} and so on up to i^{8}.

We already know that i^{1} = i and i^{2} = –1. So the rest are

i^{3} = i • i^{2} = i • -1 = -i

i^{4} = i • i^{3} = i • -i = -i^{2} = 1

i^{5} = i • i^{4} = i

i^{6} = i • i^{5} = i • i = -1

i^{7} = i • i^{6} = i • -1 = -i

i^{8} = i • i^{7} = i • -i = -i^{2} = 1

As you have probably guessed, all powers of i can be expressed in terms of just the first four powers: i, -1, -i, and 1.

Sample Problem

Solve x^{2} = -16.

It used to be that we would see a problem like this, and we would throw up our hands and say, "There's no real solution. There's nothing we can do!" and call it quits. Now that we know about i and imaginary numbers, this problem is almost too easy.

x = ± 4i

Then we double check our solutions.

(4i)^{2} = 16i^{2} = -16.

(-4i)^{2} = (-4)^{2}i^{2} = 16i^{2} = -16.

Check and check.

Complex numbers are pretty simple. They all take the form a + bi, where a and b are real numbers. So, they have both a real and imaginary component, like a baseball game between the Yankees and the X-men. Addition, subtraction, and multiplication all work pretty much how you would expect them to.

When you have an exponent of i, check to see if it 4 or less. If so, simplify. If not, find as many powers of 4 as you can, because each i^{4} equals 1, which will multiply out without consequence.

0 + 3i + 8i + 21(-1) -21 + 11i

Remember to put the real number portion first and the imaginary part second.

Example 2

Simplify.

(4i + 3)(-2i + 1)(1 – i)^{2}

Apply FOIL as needed.

(-8i^{2} + 4i – 6i + 3)(1 – 2i + i^{2})

Simplifying after each multiplication can greatly reduce the complexity of what you are working with, because each term will reduce to either a real or imaginary component.

(-8(-1) – 2i + 3)(1 – 2i + (-1))

(11 – 2i)(-2i)

-22i + 4i^{2}

-4 – 22i

Example 3

Simplify.

5i^{6} + 3 + (2i^{2} – 1)(3i – 7)

5i^{2}(i^{4}) + 3 + (2(-1) – 1)(3i – 7)

5(-1)(1) + 3 + (-2 – 1)(3i – 7)

-2 – 3(3i – 7)

You can start off by simplifying some of the imaginary terms.

-2 – 9i + 21

19 – 9i

Equations with complex numbers are often not as complex as they first appear.