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}.


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.


(-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.

green line

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



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