angle for gaussian primes ?
 

A peaceful and pleasant evening,


is it possible to calculate an angle for gaussian primes ?


for example 5=(2+i)(2-i) alpha = arc tan (2/1)


Would be nice to get a link or a clear answer,


Greetings from the gaussian primes :petrw1::uncwilly: :truck:

Bernhard

[QUOTE=bhelmes;516073]A peaceful and pleasant evening,


is it possible to calculate an angle for gaussian primes ?


for example 5=(2+i)(2-i) alpha = arc tan (2/1)


Would be nice to get a link or a clear answer,


Greetings from the gaussian primes :petrw1::uncwilly: :truck:

Bernhard[/QUOTE]

See: [url]http://mathworld.wolfram.com/ArgandDiagram.html[/url]

What's wrong with polar form?

If R = Z[i], p is a prime number, p == 1 (mod 4), then pR = PP', the product of two conjugate prime ideals. If P = (a + b*i)R, it is easily shown that the argument of a + b*i is [i]not[/i] a rational multiple of the number pi. (P[sup]k[/sup] is not a rational integer for any integer k other than 0.)

However, it is [i]also[/i] easily shown that, if p[sub]1[/sub], p[sub]2[/sub], ..., p[sub]k[/sub] are k distinct prime numbers congruent to 1 (mod 4), P[sub]j[/sub] = (a[sub]j[/sub] + i*b[sub]j[/sub])R is a prime divisor of p[sub]j[/sub]R, j = 1 to k, and

x[sub]j[/sub] = arg(a[sub]j[/sub] + i*b[sub]j[/sub])/pi <-- the circle number,

then the x[sub]j[/sub] are Z-linearly independent -- a [i]much[/i] stronger result. This result follows from unique factorization in R -- the product of integer powers of the P[sub]j[/sub] is not a rational number unless all the exponents are 0.

Why the above argument does [i]not[/i] apply to the prime divisor of 2R is left as an exercise for the reader.