# Calculators and complex numbers (Part 22)

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

$\log z = \ln r + i \theta$.

At long last, we are now in position to explain the last surprising results from the calculator video below.

Definition. Suppose that $z$ and $w$ are complex numbers so that $z \ne 0$. Then we define

$z^w = e^{w \log z}$

Naturally, this definition makes sense if $z$ and $w$ are real numbers.

For example, let’s consider the computation of $i^i$. For the base of $i$, we note that

$\log i = \log e^{\pi i/ 2} = \displaystyle \frac{\pi i}{2}$.

Therefore,

$i^i = e^{i \log i} = e^{i \pi i/2} = e^{-\pi/2}$,

which is (surprisingly) a real number.

As a second example, let’s compute $(-8)^i$. To begin,

$\log(-8) = \log \left( 8 e^{\pi i} \right) = \ln 8 + \pi i$.

Therefore,

$(-8)^i = e^{i \log(-8)}$

$= e^{i (\ln 8 + \pi i)}$

$= e^{-\pi + i \ln 8}$

$= e^{-\pi} (\cos [\ln 8] + i \sin [ \ln 8 ] )$

$= e^{-\pi} \cos (\ln 8) + i e^{-\pi} \sin (\ln 8)$

In other words, a problem like this is a Precalculus teacher’s dream come true, as it contains $e, \ln, \pi, \cos, \sin$, and $i$ in a single problem.

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