Abstract:I was taught that $e^x = x$ has no solution, and taught to leave it at that. But in mathematics "no solution" has usually meant "not on this line yet": $x^2 = -1$ waited for the complex plane, and $e^x = x$ turns out to be waiting there too. Over $\mathbb{C}$ the exponential has a fixed point $\varrho = 0.318\ldots + 1.337\ldots\,i$, the unique solution of $\exp(z) = z$ in the strip $0 < \operatorname{Im} z < \pi$ (equivalently $-W_{-1}(-1)$), and it carries more structure than its one-line definition lets on. At $\varrho$ the rectangular and log-polar coordinates of a complex number coincide, forcing the identities $\operatorname{Re}\varrho = \log|\varrho|$ and $\arg\varrho = \operatorname{Im}\varrho$. As a dynamical point $\varrho$ is repelling for $\exp$ and attracting for $\log$, linearizable for both by one Koenigs coordinate, and the base of a transpose identity $w^\varrho = \varrho^{\log w}$. It generates an aperiodic log-polar lattice and sits a hair off a clean relation with $\pi$, namely $\operatorname{Re}(\varrho)\,\pi = 0.99944\ldots$. Passing to the octonions, the fixed points of $\exp_{\mathbb{O}}$ fill concentric six-spheres, the innermost $\operatorname{Re}(\varrho) + \operatorname{Im}(\varrho)\,S^6$, whose triples obey an exact identity $I_4^2 + \tfrac14 I_5^2 = \operatorname{Gram}$ carrying one invariant, a twist angle, absent from ordinary spherical trigonometry. Throughout, what is proved is kept apart from what is only computed.
From: Michael Bickford [view email]
[v1]
Mon, 1 Jun 2026 04:22:52 UTC (34 KB)
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