# Common Core Standards: Math

#### The Standards

# 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 + 3*i* by its conjugate, 2 – 3*i*? We'll use FOIL, or double distributive, to multiply. When we distribute the 2, we get 4 – 6*i*. When we distribute the 3*i*, we get 6i – 9*i*^{2}.

So far, we're looking at 4 – 6*i* + 6*i* – 9*i*^{2}. Do you see that? The 6*i* terms cancel each other out, which leaves us with 4 – 9*i*^{2}.

Remember, since *i*^{2} = -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 *a* – *bi*), our answer will always be *a*^{2} – *b*^{2}*i*^{2}. (The two middle terms will always cancel each other out.) Since we know that the *i*^{2} part will always be -1, we can simplify it further and say that it'll always be *a*^{2} + *b*^{2}. 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 – 4*i* is 2 + 4*i*. So we'll multiply top and bottom by that. The top becomes 6(2 + 4*i*). We won't distribute just yet, since we may be able to reduce it later. The bottom becomes 2^{2} – (4*i*)^{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 *a* – *bi*, which gives us *a*^{2} – *b*^{2}*i*^{2}, or *a*^{2} + *b*^{2} since we know what *i*^{2} = -1). Then, take the square root of the answer you get. (So it means .)

Let's find the modulus of 8 – 2*i*. The conjugate of 8 – 2*i* is 8 + 2*i*. If we multiply them, we get (8 – 2*i*)(8 + 2*i*) = 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 .