\(i^i\) is punk (complex exponentiation zine)

January 2023

R., T., N., and I spent a rainy Saturday in January in a warehouse in West Oakland, at what R. called the “anarchist print shop,” making zines on their risograph. One of the zines I made was a zine version of my notes on complex exponentiation—a lucid and exciting exposition of this strange and beautiful phenomenon.

The title image for the zine I made by printing out copies of other of my math notes, ripping them up, arranging them into a tower of \(\sqrt{-1}\)s on my floor, upping the contrast, converting to black and white, and inverting. We did a print run of 50 copies. Here’s how they looked, ultimately:

Risographs are a 1970s/1980s technology that were quickly superseded by modern photocopier technology; they work similarly to mimeographs. In their afterlife they’ve become popular in hipster graphic design scenes. The anarchist print shop also has an automated folder, a vibrator, a saddle stitch, and a frightening trimmer (so frightening that to operate it, you have to hold down two separate buttons on opposite sides of the machine, as an extra safety check that your hand isn’t in its path).

“I’m pretty sure this is the first time we’ve ever printed something typeset in \(\LaTeX\),” said N., who helps run the print shop.

Here’s an excerpt from my personal notes on how it went the first time I taught this, back in 2020:

I told the kids that \(i^i\) is about \(0.2078\). Do you actually believe that? Come on, guys, do you really think that \(i^i\) is \(0.2\)ish?? Yeah, I’m just kidding! It’s actually: \[ \begin{align*} &\quad\quad\vdots\\ &31,920,519.1574213\dots,\\ &59,609.74149287215\dots,\\ &111.31777848985622\dots,\\ i^i =\quad &0.2078795763507619\dots,\\ &0.0003882032039267\dots,\\ &0.0000000724947252\dots,\\ &0.0000000001353797\dots,\\ &\quad\quad\vdots \end{align*} \]

I then asked the kids to try to calculate \(i^i\), \(\sqrt[i]{i}\), and \(2^i\). Most were able to calculate \(i^i\) (“try writing \(i\) in a different way”); of those who were, almost all were able to calculate \(\sqrt[i]{i}\) (turns out it only requires two more steps). Almost no one was able to calculate \(2^i\), which wasn’t surprising. (Doing so involved a proof technique that we had used back in the fall on our proofs of logarithm theorems, so it wasn’t out of the question, but the couple kids who were determined to figure it out needed a lot of guidance.)