```Date: Feb 15, 2014 6:52 AM
Author: g.resta@iit.cnr.it
Subject: Re: real parts of powers

On Saturday, February 15, 2014 1:01:56 AM UTC+1, quasi wrote:> Computer assistance allowed.I've used Mathematica to compute the symbolic real part of(a+b*I)^k for a few small values of k, say r(k)Then I selected 3 exponents n1, n2, n3, and I solved the system of 2equations r(n1)=r(n2)  r(n1)=r(n3) in the unknown a and b.Looking for real, nonzero solutions in a and b, solutions which shouldnot lead to powers with null Re or Im parts.I found several solutions, but often they cannot be expressed with a clean formula.3 nice examples (i've tested just a few set of exponents, I do notknow if there could exist one with rational coefficients):For n1=1, n2=2, n3=5  I got  x = -1 (+/-) I*sqrt(2) with common real part -1.For n1=2, n2=3, n3=5 I got x = -1/2 (+/-) I*sqrt(3/20)with common real part 1/10.For n1=3, n2=4, n3=6 I got  x = -1/2 (+/-) I*sqrt(6-sqrt(33))/2with common real part (17-3*sqrt(33))/8.For 4 exponents we clearly end up with 3 equations in twounknowns, so to have solutions we need the equations to be somehownot independent, which seems difficult, but I have not the slightestidea on how to prove it.g.--http://equal.to/
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