Calculators and complex numbers (Part 5)

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.

The trigonometric form of a complex number permits a geometric interpretation of multiplication, given in the following theorem.

Theorem. \left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right] = r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2]).

While this theorem doesn’t seem all that helpful — just multiplying complex numbers seems easier — this theorem will be a great help for the following problem:

Compute (\sqrt{3} + i)^{2014}. (When teaching this in class, I usually choose the exponent to be the current year.)

Let’s discuss the options for evaluating this expression.

Method #1: Multiply it out. (Students reflexively wince in pain — or knowing laughter — when I make this suggestion.)

Method #2: Use the 2014th row of Pascal’s triangle. (More pain and/or laughter.)

Method #3: Use the above theorem. It’s straightforward to write \sqrt{3} + i as 2 \displaystyle \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)… for reasons that will become apparent later, I tell my students that I’ll use radians and not degrees for this one. Most students can recognize — and this is important, before I formally prove De Moivre’s Theorem — that they need to multiply 2 by itself 2014 times and add \displaystyle \frac{\pi}{6} to itself 2014 times. Therefore,

(\sqrt{3} + i)^{2014} = \displaystyle 2^{2014} \left( \cos \frac{2014\pi}{6} + i \sin \frac{2014\pi}{6} \right) = \displaystyle 2^{2014} \left( \cos \frac{1007\pi}{3} + i \sin \frac{1007\pi}{3} \right)

I then try to coax my students to compute \displaystyle \cos \frac{1007\pi}{3} without a calculator. With some prodding, they’ll recognize that \displaystyle \frac{1007}{3} = \displaystyle {335}\frac{2}{3}, and so they can subtract 334\pi (not 335\pi) without changing the values of sine and cosine. Therefore,

(\sqrt{3} + i)^{2014} = \displaystyle 2^{2014} \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right)

= 2^{2014} \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)

= 2^{2013} (1-i\sqrt{3})

By this point, students absolutely believe that the trigonometric form of a complex number serves a useful purpose. Also, this numerical example has prepared students for the formal proof of DeMoivre’s Theorem, which will be the subject of the next two posts.

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



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