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# High School: Number and Quantity

### The Complex Number System HSN-CN.A.2

2. Use the relation i2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.

Addition and subtraction of complex numbers is easy. Just because it says "complex" doesn't mean it is. Some ice cream cartons say, "nonfat," but do you really think it doesn't have the same effect on your waistline as regular ice cream? Please.

The i looks just like a variable and in this context, it acts just like one too. The basic rule for adding and subtracting is the one your students might imagine: we add and subtract like terms. Just make sure they remember to express their answers in a + bi form. In other words, we want the i term to come after the "non-i" term.

So, for example, if we want to add 4 + 10i and 7 – 2i, we combine the like terms. That means the 4 and the 7 go together, as do the 10i and -2i. (We can do this thanks to a generous donation from the Commutative Property. Remember to send it a thank you card.) The answer, then, is 11 + 8i.

If we want to subtract, we simply remember to distribute the negative sign (à la Distributive Property).

(21 + 4i) – (16 – i) becomes 21 + 4i – 16 + i, or even more simply, 5 + 5i.

Multiplication also follows the rules of algebra. Many people use the "FOIL" method to multiply binomials (First, Outer, Inner, Last). We can also think of it as the "double distributive" property. First, distribute the real portion of the first complex number, then the imaginary part.

When multiplying, it's worth noting that i2 = -1. This will help a lot when we simplify. Don't be surprised when a real number results from us fiddling around with these imaginary numbers. It happens sometimes. We just gotta roll with it.

#### Drills

1. Find the sum of 4 + 6i and 11 – 2i.

15 + 4i

All we have to do is combine the like terms. For the real numbers, 4 + 11 = 15. On the imaginary side, 6i – 2i = 4i. Put 'em together and what do you get? Uh, well (B). It's just like second-grade math, isn't it?

2. What does (8 + 4i) – (3 + i) equal?

5 + 3i

We add and subtract numbers just like we always did—along with distributing the negative sign. When we do so, (8 + 4i) – (3 + i) becomes 8 + 4i – 3 – i. If we combine the like terms and express our answer in a + bi form, we should get (A).

3. Find the product of 6 and 3 + i.

18 + 6i

The product of 6 and 3 + i is 6(3 + i). First, we have to distribute the 6. It gets around anyway. We then have 6 × 3 = 18 and 6 × i = 6i. If we put our answer in a + bi form, like we will for now and forever, we'll get (C) as the correct answer.

4. What does (5i)2 equal?

­-25

You know that to square a monomial, you square each part of it. That means we have to consider the 5 and the i separately. We know that 52 = 25 and i2 = -1. Since the two are multiplied together, 25 × -1 = -25. In other words, (A).

5. What does (2 + i) × (3 + 4i) equal?

2 + 11i

All right! It's time for some double distributive action (or "FOIL," if that's what you prefer). First, let's distribute that 2 to the second binomial. That gives us 6 + 8i so far. Now, let's distribute that i to the second binomial. That gives us 3i + 4i2. Now, we combine them and make 6 + 8i + 3i + 4i2. But wait a minute—remember the definition of i. If i2 = -1, then 4i2 = 4(-1) or -4. That means we have 6 + 8i + 3i – 4 instead. If we simplify, our answer will be (B).

6. What is (3 – i)2?

8 – 6i

Asking us to square (3 – i) is the same as asking us to multiply (3 – i) by (3 – i). Now it's time for more double distributive. First, we distribute the 3 to the second set of parentheses to get 9 – 3i. Then, we distribute the -i to get -3i + i2. What do we know about i2, again? So that part is actually -3i – 1. If we combine the two, we get 9 – 3i – 3i – 1 and after we combine like terms (rational goes with rational, imaginary goes with imaginary) and express it in a + bi form, we get (D).

7. What does i(6 – 2i) equal?

2 + 6i

Those negative signs can get tricky, huh? Well, given i (6 – 2i), the first thing we want to do is distribute the i. That means we get 6i – 2i2. Since i2 = -1, we actually have 6i – (-2), or 6i + 2. Just switch those around so that the real number is in front, and we have (A).

8. What does 7i(7i – 5) equal?

-49 – 35i

The first thing to notice is that the stuff inside the parenthesis isn't in a + bi form. Don't let that confuse you. Proceed as normal and distribute the 7i. If we do that, we end up with 49i2 – 35i. Since we know i2 = -1, that turns into -49 -35i. That's (C)—and look! It's already in a + bi form. What could be better?

9. Solve (8 – 2i)(8 + 2i).

68

Distribute things double-time! First we distribute the 8, which gives us 64 + 16i, and then the -2i, which gives us -16i – 4i2. If we put them together, we get: 64 + 16i – 16i – 4i2. Since 16i – 16i = 0 (even in the imaginary world), we really just have 64 – 4i2. You're probably sick of hearing it, but i2 = -1, so we really have 64 + 4, which is just (D).

10. Solve (-3 + 5i)(3 – 5i).

16 + 30i

The biggest challenge with imaginary numbers and distribution is the negative signs. Here, distributing -3 gives us -9 + 15i. Then, distributing the 5i makes 15i – 25i2. That means we have -9 + 15i + 15i -25i2. Sometimes people would rather not deal with imaginary numbers (even though they're perfectly nice), so they try and cancel them out. In this case, the two 15i terms are added, not subtracted. That means we have -9 + 30i – 25i2. We don't have to tell you this is the same as -9 + 30i + 25, do we? We should get (B) as our answer.