# High School: Number and Quantity

### The Complex Number System HSN-CN.B.6

6. Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints.

The modulus of a complex number a + bi is defined as . This little (or sometimes big) number can be useful in a variety of things having to do with complex numbers.

For starters, what about the distance between two points? In geometry, the distance formula  is pretty much our Holy Grail for everything. With complex numbers, it's no different, except instead of x1 and y2, we have a1 and b2. Using these generic values, we get .

But why plug individual a and b values when we can just subtract the two complex numbers directly? If we subtract a1 + b1i from a2 + b2i, we'll get (a2 – a1) + (b2 – b2)i as our answer. Taking the modulus of that will give us .

Whoa. That means the distance between two imaginary points is the same as the modulus of the difference between the complex numbers. Pretty nifty.

Not clicking? Find the distance between 5 + 2i and 6 + 4i to help get your students' imaginary ball rolling.

Rather than plug four numbers into the distance formula, we can subtract them and reduce them to two. Solving 5 + 2i – (6 + 4i) will give us -1 – 2i.

Now, let's find that modulus. If a = -1 and b = -2, the modulus gives us a distance of . Not too shabby.

Finding the midpoint of a segment joining complex numbers is even easier. It's just like finding the midpoint of a segment joining real points. All we need to do is average the a values and average the b values.

So if we want to find the midpoint of the segment joining 3 + 8i and 5 – 2i, we would find the average of the a values:

...and find the average of the b values:

The midpoint of our line segment is 4 + 3i. Who said math was hard?

#### Drills

1. What is the modulus of 6 + 8i?

10

The modulus is the square root of the sum of a squared and b squared. In other words, it's . Since the sum of the squares is 100, the modulus is 10. Something to note: moduli will never be imaginary numbers!

2. Find the distance between 4 + 9i and 6 – 3i.

Well, we know it can't be (A), since distances are always real. First, we can perform the subtraction: 4 + 9i – 6 – 3i = ­-2 + 12i. Now, we take the modulus by taking the square root of the sum of the squares of a and b. In other words, .

3. Find the distance between 3 + 6i and 7 + 9i.

5

The first thing we need to do is calculate the difference between the complex numbers. When we subtract, we get -4 – 3i. Then, we calculate the modulus, and that will give us the distance. . So our distance is 5.

4. The distance between 9 – 2i and -7 + 3i equals the modulus of which complex number?

16 – 5i

When does the distance between two imaginary points equal the modulus a complex number? We can either calculate the modulus of each of the answer options and match it up with the distance between the points, or we could understand the relationship behind these terms. The latter sounds way easier.

The distance is the modulus of the difference between the two complex numbers. That means that our answer should equal the difference between the two complex numbers we're given. That's less work for us! The difference between 9 – 2i and -7 + 3i is the same as writing 9 – 2i – (-7 + 3i). That simplifies to 16 – 5i.

5. Find the distance between 3i and 6 + i.

Even distances between imaginary points are real distances, and we can't have negative distance. We can go toward the negative direction on a coordinate plane, but distance is never ever negative. That rules out (C).

We can subtract 6 + i and 3i, which will give us 6 ­– 2i. If we find the modulus of that number, we get . That's not an option, but only because it can be simplified to .

6. Find the distance between 9 and -13i.

This may seem tricky at first, but it's easier than you might think. Our first complex number isn't really complex since its imaginary component isn't there. That just means b = 0. Our second has no real component, which means that a = 0. So when we subtract them, we really just get 9 + 13i. Simple, right? If we find the modulus, we get .

7. Find the midpoint of the segment joining 2 + 7i and 10 – 3i.

6 + 2i

To find the midpoint, simply average the a and the b values of the given complex numbers. The average of 2 and 10 is , and the average of 7 and -3 is . That means the midpoint is 6 + 2i, or (B).

8. Find the midpoint of the segment joining 2 + 3i and 2 – 3i.

2

Once again, we average the a's and the b's to find the midpoint. We have  for our a value. Shocker. Our b values average to . There's no need to freak out. It's okay that that happened; in math, almost anything is possible. It just means that our answer is (A), since there's no i term.

9. Find the midpoint of the segment joining 3 – 11i and -1 + 5i.

1 – 3i

The midpoint just means we average the a values and the b values and make our own complex number out of them. So for our average a value, we have . Well, that eliminates (B) and (D). Now, our b values have an average of . That means (C) is the right answer.

10. If the midpoint of 2i and another point is 1 – 5i, what is the other point?