# Calculators and complex numbers (Part 6)

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 post, I used a numerical example to justify De Moivre’s Theorem:

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

The proof has two parts:

1. For $n \ge 0$: proof by induction.
2. For $n < 0$: let $n = -m$, and then use part 1.

In this post, I describe how I present part 1 to students in class. The next post will cover part 2. As noted before, I typically present this theorem and its proof after a numerical example so that students can guess the statement of the theorem on their own.

Proof for $n \ge 0$.

Base Case: $n = 0$. This is trivial, as the left-hand side is

$\left[ r (\cos \theta + i \sin \theta) \right]^0 = 1$,

while the right-hand side is

$r^0 (\cos 0 \theta + i \sin 0 \theta) = 1(\cos 0 + i \sin 0) = 1(1 + 0i) = 1$.

Assumption. We now assume, for a given integer $latex$n\$, that

$\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$.

Inductive Step. We now use the above assumption to prove the statement for $n+1$. On the board, I write the left-hand side on the top and the right-hand side on the bottom, leaving plenty of space in between:

$\left[ r (\cos \theta + i \sin \theta) \right]^{n+1}$

$\quad$

$\quad$

$\quad$

$\quad$

$\quad$

$= r^{n+1} (\cos [n+1] \theta + i \sin [n+1] \theta)$.

All we have to do is fill in the space to transform the left-hand side into the target on the right-hand side. (I like to call the right-hand side “the target,” as it suggests the direction in which the proof should aim.) I also tell the class that we’re more than two-thirds done with the proof, since we’ve finished the first two steps and have made some headway on the third. This usually produces knowing laughter since the hardest part of the proof is creatively converting the left-hand side into the target.

The first couple steps of the proof are usually clear to students:

$\left[ r (\cos \theta + i \sin \theta) \right]^{n+1} = \left[ r (\cos \theta + i \sin \theta) \right]^n \cdot \left[ r (\cos \theta + i \sin \theta) \right]$

$= \left[ r^n (\cos n \theta + i \sin n \theta) \right] \cdot \left[ r (\cos \theta + i \sin \theta) \right]$

by induction hypothesis. (I’ll also remind students that, as a general rule, when doing a proof by induction, it’s important to actually use the inductive assumption someplace.) At this point, most students want to distribute to get the right answer. This will eventually produce the correct answer using trig identities. However, I again try to encourage them to think like MIT freshmen and use previous work. After all, the distances multiply and the angles add, so the next step can be

$=r^n \cdot r (\cos [n \theta + \theta] + i \sin [n \theta + \theta])$

$= r^{n+1} (\cos [n+1]\theta + i \sin [n+1]\theta)$.

In tomorrow’s post, I’ll talk about how I present the second part of the proof to my students.

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