# A funny number

I was giving tutorials for an undergraduate course in Signals and Systems theory. Since students came from quite varying backgrounds I had to start very simple – my first tutorial used to be just on complex numbers. This left some of the students bored, saying they’ve seen it all. To keep them thinking I usually asked them to compute a complex number for me: Let $j$ be the imaginary unit, what is the value of $j^j$?

This question is not as trivial as it may seem at first sight. It requires a generalization of the exponentiation $a^b$ from $a, b \in \mathbb{R}$ to $a, b \in \mathbb{C}$. There are rigorous ways of doing this which I do not want to discuss in detail here. Let us just assume that we found a generalization of $a^b$ to complex numbers which satisfies the laws of powers, in particular the law $a^b = {\rm e}^{b \cdot \ln(a)}$ where $\ln(x)$ is the natural (base-e) logarithm. Then, we can rewrite $j^j$ as $j^j = {\rm e}^{j \cdot \ln(j)}$. What is $\ln(j)$ though? Well, $j$ can be written as ${\rm e}^{j \cdot \pi/2}$, so $\ln(j)$ should be $j \cdot \pi/2$, right? This finally brings us to $j^j = {\rm e}^{j \cdot j \cdot \pi/2} = {\rm e}^{-\pi/2}$.

Some of my students would actually obtain this result. They would usually be surprised to get a real (and sort of strange number) out of such an operation, but always they would be very proud to have the answer. The tech-savvy ones would even check their result with whatever internet-able device they carried and be extra sure to have it right.

However, my (admittedly a bit discouraging) reply would be that the answer is, despite being correct, incomplete. Actually infinitely incomplete. There are (infinitely) more “correct” answers. How so? Well, $j$ can be written as ${\rm e}^{j \cdot \pi/2}$ but it can also be written as ${\rm e}^{j \cdot (\pi/2 + 2\pi)}$ or ${\rm e}^{j \cdot (\pi/2 – 2\pi)}$, and so on. We can add any integer multiple of $2\pi$ to the phase due to the periodicity of complex numbers with respect to their phase.

For the (natural) logarithm of a complex number this means that it is in general ambiguous: $\ln(j) = j \cdot \pi/2 + 2\cdot k \cdot \pi$ for any $k \in \mathbb{Z}$. To avoid the confusion it is common to write ${\rm Ln}(x) = \ln(x) + 2\cdot k \cdot \pi$ where $\ln(x)$ is the (unique) principle value of the logarithm which we obtain by choosing the principle phase of $x$.

In that sense, the full answer would be $j^j = {\rm e}^{-\pi/2} \cdot {\rm e}^{-2 \cdot k \cdot \pi}$, where $k=0$ corresponds to the principle value of $j^j$ given by ${\rm e}^{-\pi/2}$.

Infinitely many solutions, all of them are real, and all of them connect ${\rm e}$ and $\pi$. Pretty cool, right?

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