Calculators and complex numbers (Part 9)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number z = a+bi is

z = r(\cos \theta + i \sin \theta)

where r = |z| = \sqrt{a^2 + b^2} and \tan \theta = b/a, with \theta in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that I’ll justify later in this series.

In the previous three posts, we discussed De Moivre’s Theorem:

Theorem. If n is an integer, then \left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta).

Yesterday, we used factoring to show that there are three solutions to z^3 = -27, namely, z = -3 and z = \displaystyle \frac{3}{2} \pm \frac{3\sqrt{3}}{2} i. Let’s now use De Moivre’s Theorem to take on the same task. As we’ll see, De Moivre’s Theorem provides a geometrical interpretation of this result that isn’t readily apparent using solely algebra.

Let z = r(\cos \theta + i \sin \theta), so that

z^3 = -27


r^3 (\cos 3\theta + i \sin 3 \theta) = 27 (\cos \pi + i \sin \pi)

We now match the corresponding parts. The distances from the original have to match, so that r^3 = 27, or r = 3. (Notice that there is only one answer because r must be a positive real number.) Also, the angles 3\theta and \pi must be coterminal. They do not necessarily have to be equal, but the angles must point in the same direction. Therefore,

3 \theta = \pi + 2\pi k, or \theta = \displaystyle ]frac{\pi}{3} + \frac{2\pi k}{3}

for any integer k. At first blush, it appears that there are infinitely many solutions z since there are infinitely many possible angles \theta. However, it turns out that there are only three answers, as expected.

Let’s now plug in numbers for k. Let’s use the easiest three numbers, k = 0, 1, 2.

If k = 0, then \theta = \displaystyle \frac{\pi}{3}, so that

z = 3 \displaystyle \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)

= 3 \displaystyle \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)

= \displaystyle \frac{3}{2} + \frac{3\sqrt{3}}{2} i.

If k = 1, then \theta = \displaystyle \frac{\pi}{3} + \frac{2\pi}{3} = \pi, so that

z = 3 \displaystyle \left( \cos \pi + i \sin \pi \right) = 3(-1+0i) = -3.

If k = 2, then \theta = \displaystyle \frac{\pi}{3} + \frac{4\pi}{3} = \displaystyle \frac{5\pi}{3}, so that

z = 3 \displaystyle \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right)

= 3 \displaystyle \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)

= \displaystyle \frac{3}{2} - \frac{3\sqrt{3}}{2} i.

Not surprisingly, we obtain the same three answers that we did using algebra.

What if we keep increasing the value of k? Let’s find out with k = 3:

If k = 3, then \theta = \displaystyle \frac{\pi}{3} + \frac{6\pi}{3} = \displaystyle \frac{7\pi}{3}, so that

z = 3 \displaystyle \left( \cos \frac{7\pi}{3} + i \sin \frac{7 \pi}{3} \right)

= 3 \displaystyle \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)

= \displaystyle \frac{3}{2} + \frac{3\sqrt{3}}{2} i.

In other words, since \displaystyle \frac{\pi}{3} and \displaystyle \frac{7\pi}{3} are coterminal, we end up with the same answer. This resolves the apparent paradox of having infinitely many possible angles \theta but only three solutions z.

Using De Moivre’s Theorem certainly appears to be much more difficult than just factoring! However, this solution provides a geometric interpretation of the three roots that isn’t otherwise apparent. Let’s using the trigonometric form of these three solutions to plot them in the complex plane:

complex roots

All three points lie a distance of 3 from the origin, and so they lie on the same circle. Also, the angle from the origin increases by \displaystyle \frac{2\pi}{3} as we shift from point to point. This divides the circle (with a total angle of 2\pi into three equal parts, and so the points are evenly spaced around the circle. Also, the points form the vertices of an equilateral triangle inscribed within this circle. Again, none of this would have been apparent by strictly factoring the polynomial z^3 + 27.

All of the above can be repeated for finding the roots of any equation z^n = w. This equation has n roots which lie a distance of |w|^{1/n} from the origin on the points of a regular n-gon inscribed in a circle of this radius. The only tricky part is determining the placement of the initial point (i.e., finding the initial value of \theta from which all other points can be determined.

green line

For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.



3 thoughts on “Calculators and complex numbers (Part 9)

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.