
Re: on real part of [(1+isqrt(7))/2]^n
Posted:
Feb 11, 2014 11:54 AM


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

