fibonacci like

[mike3]

...

Very nice!

I had tried it up to

\( e^{(1+i)u} - e^u = -1 \)

but gave up then...

What's the problem? This is just from setting \( r = e^u \) or \( u = \log( r ) \), since the \( \log \) needed to raise to the complex exponent is multivalued. I do not think it can be solved in closed form. The values tested were obtained by numerical root-finding methods (Newton's method, specifically). There could be some kind of infinite series formula or something, but I would not know what it is.