Applications of complex numbers include signal processing filters in engineering and vector length calculations in physics. After defining complex numbers, this lesson explores how to combine them using addition, subtraction and multiplication.

Complex Numbers

Imagine keeping track of two things at once, like eating lunch and knowing the time. This is a complex situation somewhat related to complex numbers. Another example is that a signal can have a magnitude and a phase. Complex numbers are great for describing signals. In this lesson we define complex numbers and then use math properties to add, subtract and multiply complex numbers.

Defining Complex Numbers

Indeed, a complex number really does keep track of two things at the same time. One of those things is the real part while the other is the imaginary part. For example, z = 3 + 2i is a complex number. The real part of z is 3 and the imaginary part of z is 2. The everyday meaning of ”imaginary” is something which doesn’t exist. The meaning in math is quite different. Identifying the imaginary part of a complex number is easy because it has a label. The imaginary part is the number multiplying the label i‘. That’s right, the imaginary part of 3 + 2i is the 2. Be careful because the imaginary part is not 2i. The imaginary does not include the label.

Another example? Let’s take the complex number z = -15 – 32i. Which part is real and which imaginary? The real part is -15 while the imaginary part is -32.

The great Swiss mathematician Euler invented i in 1777. The value of i is the square root of negative one. For the most part, we will use i as the label identifying the imaginary part of a complex number. Still, we may need to evaluate i2 from time to time. If i is the square root of negative one, then i2 is the square root of negative one times the square root of negative one. Thus, i2 is -1.

Section 1Properties
Lecture 1Number PropertiesFree Preview

Number Properties

Remember the commutative property? The commutative property is about ordering. When adding or multiplying, changing the order does not change the result. For example, 3 + 6i is the same as 6i + 3. How about the associative property? The associative property is about grouping. When adding or multiplying, we can group the terms in any way without changing the result. For example, (2 + 3i) + (3 – 4i) is the same as (2 + 3) + (3i – 4i) which gives us 5 – i. Lastly, do you remember the distributive property? The distributive property is about distributing a multiplication over an addition. When multiplying a number times a parenthesis containing the sum of two or more numbers, the multiplication applies to every number in the parenthesis. For example, 2(3 – 5i) is the same as 2(3) + 2(-5i) which gives us 6 – 10i.

In the following examples we will use these four complex numbers:

  • z1 = 2 + 3i
  • z2 = -3 + 2i
  • z3 = 4 – 2i
  • z4 = -2 – 4i

Adding Complex Numbers

Example: Add z1 to itself

Start by substituting:

z1+z1

The associative property allows any grouping, so we can eliminate the parentheses:

associative

The commutative property allows any ordering for adding. The goal is to add the real parts and imaginary parts separately:

commutative

Adding the real parts gives us 2 + 2 = 4. Adding the imaginary parts gives 3 + 3 = 6. The answer is 4 + 6i.

Example: Calculate z2 + z3 + z4.

First, substitute:

z2+z3+z4

Then, we can remove the parentheses (associative property):

associative

Did you see where the +(-2 – 4i) became -2 – 4i ? We can think of +(-2 – 4i) as +1 times (-2 – 4i). The +1 times the (-2 – 4i) gives -2 – 4i.

Now, we can reorder the addition (commutative property):

commutative

Adding the real parts gives -3 + 4 – 2 = -1. Adding the imaginary parts, 2 – 2 – 4 = -4. The answer is -1 –

Lecture 2How to Solve Quadratics with Complex Numbers as the SolutionFree Preview
When you solve a quadratic equation with the quadratic formula and get a negative on the inside of the square root, what do you do? The short answer is that you use an imaginary number. For the longer, more helpful answer, check out this lesson!

Imaginary & Complex Numbers

By allowing ourselves to imagine that square roots of negative numbers actually exist, we are able to solve a lot of real-world problems. So, what does that actually look like? How would you go about solving a problem that has an imaginary solution? That’s what this lesson is all about.

You’ll need to remember that imaginary numbers come about when we take the square root of a negative number, and complex numbers are when we combine a real number with an imaginary one using addition or subtraction. If this is a new concept for you, then you should check out the previous lesson that introduces these ideas, but if you already know this, we’re ready to look at the example ‘find the roots of y = 2x^2 – 5x + 7.’

The quadratic formula
Quadratic Formula 1

Quadratics with Complex Solutions

When a problem asks you for the roots, it is the same thing as asking for the zeros or the x-intercepts. These are the points where y = 0, so we can substitute that value in to begin with.

Given the fact that we’re doing this example in the lesson all about complex number solutions, there’s a good chance that’s what this problem will have. But how could we know this if this problem wasn’t in this context?

It’s all going to come down to the discriminant. Given a quadratic equation in standard form (y = ax^2 + bx + c), the discriminant is b^2 – 4ac. If the discriminant is positive, you’ll get two real answers. If it’s equal to zero, you’re only going to get one real answer. But if the discriminant is negative, that’s when you get two complex solutions to your problem.

This is the case because of the way the quadratic formula works. If we begin to solve the example using the quadratic formula, as shown here, you might notice that the discriminant is the part of the formula that is inside the square root. Therefore, it makes sense that if that part is negative, we’ll get an imaginary number there, making our answer a complex number.

So, back to our example, if we substitute a, b and c into the formula and then begin to evaluate the expression, we’ll find the discriminant will, sure enough, become negative. You’ll see that we end up with the square root of -31, which is going to be an imaginary number. While this is what makes this problem new and different, the only special thing to do is to put an i in front of the square root to indicate it’s imaginary, make everything else on the inside of the square root positive and then continue on just like we normally would. That’s it! The i tells everyone that it is an imaginary number, but then we can move on and continue to solve the problem as normal.

As it turns out, there isn’t even anything else to do with this problem; it’s as simplified as it can be. Maybe if we could simplify the square root somehow, we might have a few more steps, but we can’t do that, so we’re good to go!

The simplified answer for the quadratic example after putting an i in front of the square root
quadratics with complex solutions

Polynomials with Complex Solutions

We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. Take this example: Solve 0 = (x – 9)^2 * (x^2 + 9).

This isn’t a quadratic equation anymore because there are two x^2s (here and here). We don’t have any fancy formula for such problems like we do for quadratics, but because of the way this one is written, we can still solve it.

We’re going to use the zero product property for this one. That’s the property that says anytime you multiply two things together and get zero, one of the things you multiplied in the beginning must have been zero. In this problem, that means that either (x – 9)^2 or (x^2 + 9) must be zero. Now that we’ve split the equation up, we’ve got two smaller equations that we do know how to solve simply with inverse operations.

Getting the x by itself in this first one means undoing a power of 2 with a square root. The square root of zero is still just zero, so that leaves us here. Now, undoing -9 with +9 tells us that x = 9.

 
Section 2