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Timeline for Do complex numbers really exist?

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Sep 17 at 16:08 comment added Thomas Andrews Yeah, and the general Taylor series around real $a$ of that function gives a radius of convergence $\sqrt{a^2+1},$ which really hints at the geometric nature of what is "obstructing" the power series... @M.Winter
Sep 17 at 15:06 comment added M. Winter The fact that the Taylor series of $(x^2+1)^{-1}$ has convergence radius 1 is one of the best example for how complex-valued reasoning can be forced on you even though both the problem and the solution are real-valued. I find this more convincing than computing roots of cubics because for those one might object that Cardano maybe hasn't found the "right" equation for the roots.
S Mar 21, 2012 at 5:30 history answered Thomas Andrews CC BY-SA 3.0
S Mar 21, 2012 at 5:30 history made wiki Post Made Community Wiki by Thomas Andrews