giovanni wrote: >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.
That was my first solution.
Note that (n1,n2,n3) = (1,2,5) is the lexicographically least triple for which a solution is possible. > >For n1=2, n2=3, n3=5 I got x = -1/2 (+/-) I*sqrt(3/20) >with common real part 1/10.
I had the above solution as well.
>For n1=3, n2=4, n3=6 I got x = -1/2 (+/-) I*sqrt(6-sqrt(33))/2 >with common real part (17-3*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.