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

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