Calculators and complex numbers (Part 12)

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, we made the following definition for $z^q$ if $q$ is a rational number and $-\pi < \theta \le \pi$ . (Technically, this is the definition for the principal root.)

Definition. $z^q = r^q (\cos q \theta + i \sin q \theta)$ .

As it turns out, one of the usual Laws of Exponents remains true even if complex numbers are permitted.

Theorem. $z^{q_1} z^{q_2} = z^{q_1 + q_2}$

Proof. Using the rule for multiplying complex numbers that are in trigonometric form:

$z^{q_1} z^{q_2} = r^{q_1} (\cos q_1 \theta + i \sin q_1 \theta) \cdot r^{q_2} (\cos q_2\theta + i \sin q_2 \theta)$

$= r^{q_1+q_2} ( \cos [q_1 \theta +q_2\theta] + i \sin [q_1\theta +q_2 \theta])$

$= r^{q_1+q_2} ( \cos [q_1+q_2]\theta + i \sin [q_1+q_2] \theta)$

$= z^{q_1+q_2}$

However, other Laws of Exponents no longer are true. For example, it may not be true that $(zw)^q$ is equal to $z^q w^q$ . My experience is that this next example is typically presented in secondary schools at about the time that the number $i$ is first introduced. Let $z = -2$ , $w = -3$ , and $q = 1/2$ . Then

$\sqrt{-2} \cdot \sqrt{-3} = i \sqrt{2} i \sqrt{3} = -\sqrt{6} \ne \sqrt{6} = \sqrt{(-2) \cdot (-3)}$ .

Furthermore, the expression $(z^{q_1})^{q_2}$ does not have to equal $z^{q_1 q_2}$ if $z$ is complex. Let $z = -1$ , $q_1 = 3$ , and $q_2 = 1/2$ . Then

$\left[ (-1)^3 \right]^{1/2} = (-1)^{1/2}$

$= [1 (\cos \pi + i \sin \pi)]^{1/2}$

$= \displaystyle 1^{1/2} \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)$

$= 1(0+1i)$

$= i$ .

However,

$(-1)^{3/2} = [1 (\cos \pi + i \sin \pi)]^{3/2}$

$= \displaystyle 1^{3/2} \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right)$

$= 1(0-1i)$

$= -i$ .

All this to say, the usual Laws of Exponents that work for real exponents and positive bases don’t have to work if the base is permitted to be complex… or even negative.

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

Calculators and complex numbers (Part 12)

Published by John Quintanilla

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