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Topic: on real part of [(1+isqrt(7))/2]^n
Replies: 20   Last Post: Feb 15, 2014 5:52 AM

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Karl-Olav Nyberg

Posts: 399
Registered: 12/6/04
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 11, 2014 11:54 AM
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On Tuesday, February 11, 2014 5:23:50 PM UTC+1, Robin wrote:
> On 11/02/2014 16:12, konyberg wrote:
>
>
>

> > a_n = Re(a^n)
>
> > a^n = (x + iy)^n = r^n (cos(nu) + isin(nu))
>
> > a_n = r^n cos(nu)
>
> > a = (1 + isqrt(7))/2 gives r = sqrt(2), and u = pi/4 (45 deg)
>
>
>
> Oh no, u = tan^{-1}(sqrt(7)) which is not a rational multiple of pi.


Hi.

Hm. I try this again. You might be correct :)

a^1 = (1 + i*sqrt(7))/2
r = sqrt(2)
a_1 = 1/2 (the error I made)
sqrt(2)*cos(1*u) = 1/2
cos(u) = sqrt(2)/4. You are correct! Thank you! (cos^(-1)(sqrt(2)/4) = tan^(-1)(sqrt(7))

However. I dont think this changes the conclusion.

KON



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