# Calculators and complex numbers (Part 11)

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.

In today’s post, at long last, I can explain one of the unexpected results of the calculator shown in the opening sections of the video below: the different answers for $(-8)^{1/3}$ and $(-8+0i)^{1/3}$.

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 previous posts, we discussed De Moivre’s Theorem:

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

This motivates the following definition:

Definition. If $q$ is a rational number, then $z^q = r^q (\cos q \theta + i \sin q \theta)$ if $theta$ is chosen to be in the interval $-\pi < \theta \le \pi$.

Technically speaking, this defines the principal value of $z^q$; however, for the purposes of this post, I’ll avoid discussion of branch cuts and other similar concepts from complex analysis. When presenting this to my future secondary teachers, I’ll often break the presentation by asking my students why it’s always possible to choose the angle $\theta$ to be in the range $(-\pi,\pi]$, and why it’s necessary to include exactly one of the two endpoints of this interval. I’ll also point out that this interval really could have been $[0,2\pi)$ or any other interval with length $2\pi$, but we choose $(-\pi,\pi]$ for a very simple reason: tradition.

Using this definition, let’s compute $(-8+0i)^{1/3}$. To begin,

$-8+0i = 8 (\cos \pi + i \sin \pi)$.

So, by definition,

$(-8+0i)^{1/3} = 8^{1/3} \displaystyle \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$

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

$= 1 + i \sqrt{3}$

As noted in an earlier post in this series, this is one of the three solutions of the equation $z^3 = -8$. Using De Moivre’s Theorem, the other two solutions are $z = -2$ and $z = 1 - i \sqrt{3}$.

So, when $(-8+0i)^{1/3}$ is entered into the calculator, the answer $1 + i \sqrt{3}$ is returned.

On the other hand, when $(-8)^{1/3}$ is entered into the calculator, the calculator determines the solution that is a real number (if possible). So the calculator returns $-2$ and not $1 + i \sqrt{3}$.

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

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