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Topic: real parts of powers
Replies: 9   Last Post: Feb 15, 2014 1:41 PM

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quasi

Posts: 10,208
Registered: 7/15/05
Re: real parts of powers
Posted: Feb 15, 2014 1:41 PM
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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):


Good question.

>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,


Right.

>but I have not the slightest idea on how to prove it.

Me neither.

quasi



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