# Calculators and complex numbers (Part 24)

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 $z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

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.

Theorem. If $z = x + i y$, where $x$ and $y$ are real numbers, then $e^z = e^x (\cos y + i \sin y)$

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$. Then we define $\log z = \ln r + i \theta$.

Definition. Let $z$ and $w$ be complex numbers so that $z \ne 0$. Then we define $z^w = e^{w \log z}$

Technical point: for the latter two definitions, these are the principal values of the functions. In complex analysis, these are usually considered multiply-defined functions. But I’m not going to worry about this technicality here and will only consider the principal values.

This is the last post in this series, where I state some generalizations of the Laws of Exponents for complex numbers.

In yesterday’s post, we saw that $z^{w_1} z^{w_2} = z^{w_1 + w_2}$ as long as $z \ne 0$. This prevents something like $0^4 \cdot 0^{-3} = 0^1$, since $0^{-3}$ is undefined.

Theorem. Let $z \in \mathbb{C} \setminus \{ 0 \}$, $w \in \mathbb{C}$, and $n \in \mathbb{Z}$. Then $(z^w)^n = z^{wn}$.

As we saw in a previous post, the conclusion could be incorrect outside of the above hypothesis, as $\displaystyle \left[ (-1)^3 \right]^{1/2} \ne (-1)^{3/2}$.

Theorem. Let $u \in \mathbb{R}$ and $z \in \mathbb{C}$. Then $(e^u)^z = e^{uz}$.

Theorem. Let $x, y > 0$ be real numbers and $z \in \mathbb{C}$. Then $x^z y^z = (xy)^z$.

Again, the conclusion of the above theorem could be incorrect outside of these hypothesis, as $(-2)^{1/2} (-3)^{1/2} \ne \left[ (-2) \cdot (-3) \right]^{1/2}$. For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

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