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

 Messages: [ Previous | Next ]
 g.resta@iit.cnr.it Posts: 43 Registered: 11/26/09
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(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.

g.
--
http://equal.to/

Date Subject Author
2/14/14 quasi
2/14/14 g.resta@iit.cnr.it
2/14/14 quasi
2/14/14 Tucsondrew@me.com
2/14/14 quasi
2/14/14 quasi
2/14/14 quasi
2/14/14 quasi
2/15/14 g.resta@iit.cnr.it
2/15/14 quasi