# High School: Number and Quantity

### The Complex Number System HSN-CN.A.3

3. Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers.

Every complex number has an important partner: its conjugate. (Whether they're business partners, square-dancing partners, or domestic partners, we don't really know.)

To find the conjugate of a complex number, your students should know to change the sign between the real and the imaginary part. Or, in other words, change the sign in the middle so that a + bi becomes a – bi.

A curious thing happens when we multiply conjugates: the product becomes real. What happens when we multiply 2 + 3i by its conjugate, 2 – 3i? We'll use FOIL, or double distributive, to multiply. When we distribute the 2, we get 4 – 6i. When we distribute the 3i, we get 6i – 9i2.

So far, we're looking at 4 – 6i + 6i – 9i2. Do you see that? The 6i terms cancel each other out, which leaves us with 4 – 9i2.

Remember, since i2 = -1, that last term is 9 (not -9!). So our answer is 4 + 9, or 13.

Out of the land of the imaginary came a very real result. Pretty cool, right?

You know what's even cooler? That will always happen when we multiply conjugates! (It'll work just as well for you or your students or your 103-year-old grandmother as it did for us.)

When we multiply a pair of conjugates (say, a + bi and abi), our answer will always be a2b2i2. (The two middle terms will always cancel each other out.) Since we know that the i2 part will always be -1, we can simplify it further and say that it'll always be a2 + b2. Ain't that a nifty little shortcut?

When imaginary numbers get too meta for us, we can use conjugates as a way to turn them into real numbers.

We covered addition, subtraction, and multiplication. We even threw a few exponents up in there. What about division?

Remember the rule that says you can't leave a radical in the bottom of a fraction? (Yeah, it never really made sense to us either—why the top but not the bottom? But it's kind of like questioning why green lights mean, "Go" while purple lights don't mean anything.)

Anyway, that same rule applies to terms containing i, since i is defined as the square root of -1. To rationalize complex denominators, simply multiply the top and bottom of the fraction by the conjugate. (Yes, we did say "simply." Do it enough times, use our patented shortcut, and it will be simple.)

Let's take the fraction

The conjugate of 2 – 4i is 2 + 4i. So we'll multiply top and bottom by that. The top becomes 6(2 + 4i). We won't distribute just yet, since we may be able to reduce it later. The bottom becomes 22 – (4i)2, or 4 + 16. That's just 20.

We can factor out another 2 from inside the parenthesis.

And then simplify it a bit.

or

It really doesn't matter which one we use. They're both the same.

If we had a binomial on top, there'd be a little more algebra involved—frequently some FOIL.

Okay, so we can find conjugates by switching the signs. We can find quotients by using the conjugates. What's left?

We're glad you asked. One more thing we can do with complex conjugates: find their moduli. The modulus of a complex number is the square root of the product of a complex number and it's conjugate. Got that?

We didn't think so. We'll repeat it. The modulus of a complex number is the square root of the product of a complex number and it's conjugate.

In English, all it means is this: Multiply a complex number by its conjugate. (That means multiply a + bi by abi, which gives us a2b2i2, or a2 + b2 since we know what i2 = -1). Then, take the square root of the answer you get. (So it means .)

Let's find the modulus of 8 – 2i. The conjugate of 8 – 2i is 8 + 2i. If we multiply them, we get (8 – 2i)(8 + 2i) = 64 + 4 = 68. To find the modulus, we take the square root of 68, which we'll want to simplify. We can split 68 into 2 × 2 × 17. That means we can reduce  to .

#### Drills

1. Find the conjugate of 5 – 3i.

5 + 3i

To find the conjugate of a complex number, we only have to change the sign in the middle. That means that if we have 5 – 3i, all we do is change that – sign to a + sign. Seriously. That's all it takes. So our answer is (D).

2. Find the product of 4 – i and its conjugate.

17

The conjugate of 4 - ­– i is 4 + i. We can use the shortcut (a + bi)(abi) = a^2 + b^2 to find their product. In this case, we have a = 4 and b = 1, so our answer is 4^2 + 1^2 = 16 + 1 = 17. That's (A).

3. Find the quotient .

Since there aren't any imaginary numbers in the denominator, we don't need to use any conjugates to simplify the fraction. That's a relief, but don't go jumping to (A) just yet.

We can still reduce the numbers themselves. If we factor out a 3 from the numerator, we can simplify the fraction to . That's (B), and that's the right answer.

4. Find the quotient .

Imaginary numbers in the denominator? That's a big no-no. The best way to get rid of imaginary number is to make them real using their conjugates. The conjugate of the denominator is 8 + 6i. We should multiply the top and the bottom by that. The top becomes 10(8 + 6i) and the bottom becomes 82 + 62 = 64 + 36 = 100. Putting them together, we have , or . But we're not done there. We can reduce it even further, right? Pulling out a 2 from the top, we get .

5. Find the quotient

The first step here is to disregard the numerator. We need the denominator to not have an i, so it's conjugate time. If we multiply the bottom by 2 + 5i, its conjugate, we'll get 22 + 52 = 4 + 25 = 29. Just knowing that, (A) and (C) already look like better picks.

If we multiply 2 + 5i by 7i, we'll have 14i + 35i2, or 14i – 35. In a + bi form, that's -35 + 14i. Since we can't simplify anything else, our complete fraction looks like (C).

6. Find the quotient

To solve the problem, all we need to do is multiply the numerator and denominator by the conjugate of the denominator, 4 – 3i. The bottom would give us 42 + 32 = 16 + 9 = 25. For the top, we need to distribute 5i to each term in the conjugate, but maybe we should hold off on that.

Since we now have , we can cross out the 5 on top and reduce the 25 to 5. That will give us . Now, we can distribute the i to get 4i – 3i2, or 3 + 4i. Looking at the numerator and denominator, the only answer that works is (C).

7. Find the quotient

The fact that we have two imaginary numbers doesn't change anything except the algebra. We still have to multiply the top and the bottom by the conjugate of the denominator. It just happens to be the numerator in this case. Yes, it does mean we square the top. We can handle that.

That means we get . The top becomes 4 + 2i + 2i + i2, or 3 + 4i. The bottom becomes 5 (since 22 + 12 = 4 + 1 = 5, yes?). Put one on top of the other, and we end up with (C) as our answer.

8. Find the quotient

Multiplying the top and the bottom by the denominator's conjugate (which is 5 – 2i) will give us (3 – i)(5 – 2i) for the top and 52 + 22 = for the bottom. If we distribute the top (either by double distributive or FOIL), we'll get 15 – 5i – 6i + 2i2. That translates to 15 – 2 – 5i – 6i, or 13 – 11i. So our final answer is (A).

9. Find the modulus of 5 + 3i.

To find the modulus, need to multiply the number by its conjugate and take the square root. Let's start from step one: we first need to multiply the 5 + 3i by its conjugate.

Sure, (A) is its conjugate, but it isn't the modulus. If we find the product, we should get 52 + 32 = 25 + 9 = 34. Sure, that's (B), but we haven't taken the square root yet. Clearly, these answers don't want us to actually finish the problem.

If we take the square root of 34, we get  There's no way to simplify it since its two factors are 2 and 17. That means (D) is our answer.

10. Find the modulus of -3 + 8i.