# High School: Number and Quantity

### The Complex Number System HSN-CN.B.5

5. Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation. For example,  because  has a modulus 2 and argument 120°.

As the name suggests, complex numbers can occasionally get a little, well, complex. But relax. We won't drop complex bombs on you right now.

As we know, the x-coordinate of a complex number represents its real part, and the y-coordinate represents its imaginary part. So adding and subtracting complex numbers is pretty much combining like terms. Work first with the real component, a. Then, move on to the b, the imaginary component. Put them together what do you get? Well, uh, the answer.

For instance, let's say we have to graph the point made by adding 3 + 2i and 6 – i. We should add the a values first. Going over 3, then over 6 more, and we get to 9. So the a value is 9.

Now the b values. First, we go up 2 units (because of 2i), but then back down one for that -i term. That puts us at i. That means our point is at 9 units to the right and one i unit up. That's because 3 + 2i + 6 – i = 9 + i.

Easier than balancing a walrus on your head, right? Hopefully. Ready for that complex part? Neither are we.

Multiplication of imaginary numbers in a + bi form is easy. Students can use FOIL as though the i were an x or some other variable. But when we switch to polar coordinates, things get a little more… challenging.

Here's the basic rule to find the product of two complex numbers in polar form:

So, to find the product of (4, 30°) and (7, 20°), we just multiply 4 and 7 for the radial coordinate, and add 30° and 20° for the angular coordinate. Our product is (28, 50°). That's not hard at all, right?

Finally, we should cover how to find the reciprocal of a complex number. If it's in a + bi form, it's just algebra. Put it under 1, then multiply the top and bottom by the conjugate of the bottom.

What we mean is that the reciprocal of a + bi is . Since we can't have imaginary numbers in denominators, we have to multiply by the conjugate, which will give us , or .

Sometimes, it will involve using FOIL or the double distributive property on either the top or the bottom (or both). It's just algebra, but it can sometimes be a substantial dose of it.

Students should know that in polar form, the reciprocal of the number (r, θ) is 1 over the r value and the negative angle. For instance, the reciprocal of (2, 30°) is (½, -30°). More generally, the reciprocal of (r, θ) is (1r, -θ).

That's just one of the reasons these numbers are called complex!

#### Drills

1. What does 11 – 2i – (9 – 5i) equal?

2 + 3i

Whenever we have a minus sign outside parentheses, we have no choice. Distributing the negative is mandatory. That gives us 11 – 2i – 9 + 5i. All we need to do is combine the like terms, and the result is 2 + 3i.

2. Add 2 – i and 6 + i.

8

When we combine like terms, the imaginary terms cancel each other out. So we can add two complex numbers and end up with a real number. It's not only possible, it's not unusual, either. Regardless, our answer is (D).

3. Which of the following is the sum of the two points above?

9 + 4i

The first point is three units to the right and one unit down, which equates to 3 – i. The second point is six units to the right and five units up, which means 6 + 5i. If we add the two together, we get 9 + 4i, which is (C).

4. Find the product of 2 – 3i and 5 + 2i.

16 + 11i

The product of two binomials is really just the double distributive property, sometimes known as FOIL. To do that, we distribute the 2 to get 10 + 4i. Then, we distribute the -3i to get -15i – 6i^2. Combining the two, we get 10 + 4i – 15i + 6 (since i^2 = -1), which gives us (A).

5. Which of the following is the product of the two points above?

8 + 12i

First, we should figure out the complex numbers that these points represent. The first is two units to the left and two units up, which means -2 + 2i. The second is one unit to the right and five units down, which equals 1 – 5i. Now that we have the points, we can multiply them.

The FOIL and double distribution methods should give us -2 + 10i + 2i + 10, which reduces down to 8 + 12i, which is (B).

6. The two points on the graph represent two complex numbers. The product of these two complex numbers can also be plotted on such a graph. Which of the following graphs plots this point?

Before we can multiply anything, we need to translate the points on the graph to complex numbers. If we do that correctly, we should end up with -3i and 2 + i. We can multiply those numbers together without FOIL (aluminum or otherwise).

When we do, we get 3 – 6i. How does that help us when all our answer choices are graphs? Well, we can translate 3 – 6i to be a point on a graph. What we do is move three units to the right and six units down. If we do that, we'll see that (A) is the right answer.

7. Find the product of (3, 20°) and (9, 65°).

(27, 85°)

The rules for multiplying polar coordinates are very simple: multiply the radii and add the angles. Multiplying 3 × 9 gives us 27, and 20° + 65° = 85°. That's all it takes.

8. Find the product of (1, 22°) and (7, 0°).

(7, 22°)

Multiplying the radial coordinates together gives us 7. The rules of multiplication haven't changed because we're working with complex number; 1 times 7 is still 7. We have to add the angles though, which gives us 22° + 0° = 22°. That means (B) is the right answer.

9. Find the reciprocal of (3, 26°).

The rules for reciprocals are a bit twisted for polar coordinates. We take the r value and put it under 1, but we want the negative of the angle. So in this case, we would change 3 to ⅓ and put a negative sign in front of 26°. The answer that has both of those is (D).

10. Find the reciprocal of .