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}.

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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