Calculators and complex numbers (Part 21)

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.

Over the past few posts, we developed the following theorem for computing e^z in the case that z is a complex number.

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.

Of course, this looks like what the definition ought to be if one formally applies the Laws of Logarithms to r e^{i \theta}. However, this complex logarithm doesn’t always work the way you’d think it work. For example,

\log \left(e^{2 \pi i} \right) = \log (\cos 2\pi + i \sin 2\pi) = \log 1 = \ln 1 = 0 \ne 2\pi i.

This is analogous to another situation when an inverse function is defined using a restricted domain, like

\sqrt{ (-3)^2 } = \sqrt{9} = 3 \ne -3


\sin^{-1} (\sin \pi) = \sin^{-1} 0 = 0 \ne \pi.

The Laws of Logarithms also may not work when nonpositive numbers are used. For example,

\log \left[ (-1) \cdot (-1) \right] = \log 1 = 0,


\log(-1) + \log(-1) = \log \left( e^{\pi i} \right) + \log \left( e^{\pi i} \right) = \pi i + \pi i = 2\pi i.

green line

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



3 thoughts on “Calculators and complex numbers (Part 21)

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