Inverse Functions: Logarithms and Complex Numbers (Part 30)

Ordinarily, there are no great difficulties with logarithms as we’ve seen with the inverse trigonometric functions. That’s because the graph of y = a^x satisfies the horizontal line test for any 0 < a < 1 or a > 1. For example,

e^x = 5 \Longrightarrow x = \ln 5,

and we don’t have to worry about “other” solutions.

However, this goes out the window if we consider logarithms with complex numbers. 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. 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

This material appeared in my previous series concerning calculators and complex numbers:




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