High School: Number and Quantity

The Complex Number System HSN-CN.B.4

4. Represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number.

Up until now, the odds are pretty good that all your students' graphing has been on the Cartesian plane. (Their graph paper has always looked like a grid of little blue squares on a white background, with the x and y axes representing real numbers.)

We know that any point in a plane can be represented in terms of an x and a y coordinate. Those coordinates are real. Well, that's all over. We're going to change the types of numbers and then change the whole graph.

Instead of the y-axis representing real numbers, it's going to represent imaginary numbers. In other words, we can label it i, 2i, 3i, etc., going up from the origin, and -i, -2i, ­-3i, etc. going down.

Adding i's to the y-axis has made it possible to take any complex number in a + bi form and to locate it on the plane. The x-coordinate is the "a" and the y-coordinate covers the "bi" part. Tell your students to hold their applause, please.

So, to plot 3 + 7i, we start at the origin. We go over 3 units on the x-axis, and up 7i units on the y-axis. Our point would be in the first quadrant, right here.

If our point was -4 – 11i, we would go 4 to the left and 11i down, and plot the point in the third quadrant. (And yes, we do expect students to know the quadrants: they start with I in the top right and go counterclockwise.)

But, wait, it gets even better. Think back to every single corny airplane movie you've ever seen. You know the scene where the air traffic controller realizes there's a problem, and they show his screen as he's running around the room in a panic? Think of his screen. There were no tiny little blue boxes on a white background. No, his graph was cool. It wasn't at all square; it was circular.

The different dots on this screen are each represented by two things: an angle and a radius. The coordinates are (r, θ). The coordinate r is the distance away from the pole. (Not the North Pole. The "pole" is essentially the same as the origin.) The coordinate θ is the angle made by the point compared to a horizontal line.

These are called polar coordinates and have no relation to polar bears. Complex numbers can be represented using polar coordinates (not so much polar bears).

We won't lie to you. The conversion process from Cartesian to polar might look kind of scary until your students get the hang of it. Don't despair, though. It's not easy, but it's not rocket science, either (although they are applicable to airplanes, so…).

The first step is to find the magnitude (its distance from the pole, or origin). The magnitude formula is a take-off from our good old distance formula.

In our example, the magnitude would be , or , or 5. We call this coordinate r, for radius. For the record, they don't always turn out to be whole numbers. We told you 4 + 3i looked friendly, didn't we?

Okay, great. So we have our first coordinate. We know how far out the number is. Now we just need to find the angle that would get the direction down. How on Earth do we do that?

To find the angle in question, θ, we use the following formula:

In the case of 4 + 3i, we get:

We'll let our calculator do its magic. If it's being cooperative, we should get θ ≈ 37 degrees.

So, our rectangular coordinates of 3 + 4i become (5, 37°) when translated to polar coordinates. Piece of (airplane-food) cake.

Drills

1. In which quadrant would we find the point 2 – 6i?

IV

Since our a value is 2, we go over 2 units to the right on the x-axis. The b value of -6i means we go down 6i. Since the quadrants start in the top right and go counterclockwise, that places us in the fourth quadrant. Hopefully you know Roman numerals enough to say that (D) is the right answer.

2. In what quadrant would we find the point -6 + i?

II

The value for a is -6 and the value for b is 1. So we go 6 units to the left of the origin (left because the a value is negative), and up 1 unit. We're in the negative real numbers and positive imaginary numbers, so we're in the second quadrant.

3. In what quadrant would we find the point -1 – 2i?

III

Both our a and b values are negative. That means we go to the left and down from the origin. Since we know that the quadrants are numbered starting with the top right as I and going counterclockwise, out point is in the third quadrant.

4. Is it possible to graph points without an a value? How do we plot 7i?

Go up 7 units and stay on the y-axis

Okay, we admit it. This was kind of a trick question. The a value still exists; it's just 0. (Though to be fair, it wasn't all that hard!) So you go over 0 and up 7. That's all there is to it.

5. Convert 4 + 4i to polar coordinates.

First, we find the r coordinate. We can do that using the magnitude formula . If we substitute in our values, we get . What about θ? Well, we just take the inverse tangent of ba. (That's what the formula says, right?) That translates to arctan (44) = arctan(1) = 45. That means θ = 45°.

6. Convert 5 – i to polar coordinates.

First, we find r. Plugging in 5 for a and -1 for b, we get . Unfortunately, there's no way to make that any simpler. Now for θ. Using those same a and b values, we should get the inverse tangent of -⅕. With the help of our handy dandy calculator, we can find arctan(-⅕), which is approximately 11°.

7. Convert -6i to polar coordinates.

(6, -90°)

For the r value, we can easily see why it would be 6. After all, . That's not really the issue here, except for (C).

Calculating θ might be a bit problematic, since a denominator of 0 for the inverse tangent value might blow up the world. (Remember that dividing by zero is an offense punishable by death—or worse: more math problems.) We should never forget common sense, though.

If we assume our Cartesian imaginary graph, the -6i coordinate would be like going down the y-axis. Compared to a horizontal, that's 270° or -90° (since we're going down 90° rather than up). That gives us our answer right there, and we didn't have to divide by zero!

8. Which of the following complex numbers is equivalent to ?

5 + 3i

Yes, we do have to calculate the polar coordinate of all the answer choices. At least (C) can be eliminated just by calculating r (). Then, we have to take a look at the inverse tangents. Remember, the fraction is ba, and not the other way around. If we calculate the value of θ for (A), (B), and (D), we end up with (A) as the right answer.

9. Which of the following complex numbers is equivalent to ?

2 + 11i

The best way to quickly screen these polar coordinates is by calculating the r coordinate. Only values of 11 and 2 will give us . That eliminates (A) and (D). It all comes down to θ. The difference between (B) and (C) is a matter of 112 or 211. Since the inverse tangent of 112 gives us the 80° we want, our answer is (B).

10. What do the two coordinates r and θ represent in a polar graph?