I mean the opening of a Cartesian system, and then the natural logarithm. Later there were complex numbers.

]]>Dear Lei, here is a direct argument that you asked for in our email correspondence. Assume that . Following the blog entry, we can assume that has no spectrum at , for otherwise . Then,

,

hence . Furthermore, the sign of equals the sign of , which in turn equals depending whether lies in the component or the component. Done.

]]>therefore ; we have.

$latex e^{\frac{1+i \sqrt{n}}{n}} = e^{\frac{1}{n}}

]]>I think you are correct. In particular, is always singular on the components. It would be interesting to have an elementary proof of this in the case when n is even (the argument I gave only worked then n is odd). It generalises the assertion that an orthogonal matrix of determinant -1 necessarily has an odd number of eigenvalues at -1 (because the product of all the other eigenvalues is positive), regardless of the parity of dimension.

]]>I believe this is the case when is odd. For instance, for odd , the component of containing will necessarily have an odd number of eigenvalues at -1 (counted with algebraic multiplicity), since all nearby eigenvalues come in pairs , and so is necessarily zero on this component. (One way to think about this is that one inequality comes from considering the projection of onto the totally positive space, and the other inequality comes from analysing the totally negative space. When is even, it looks like both arguments give the same inequality rather than opposing inequalities.)

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