
Re: real parts of powers
Posted:
Feb 15, 2014 6:52 AM


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 2 equations 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 should not 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 not know 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(6sqrt(33))/2 with common real part (173*sqrt(33))/8.
For 4 exponents we clearly end up with 3 equations in two unknowns, so to have solutions we need the equations to be somehow not independent, which seems difficult, but I have not the slightest idea on how to prove it.
g.  http://equal.to/

