if you look at the wolframalpha calculation page:
-13-26i - Wolfram|Alpha
it factorises -13-26i as
i×(1+2 i)×(2+3 i)×(3+2 i)
my question is why is the i a necessary factor to take out? are the rules regarding this?
Thanks
if you look at the wolframalpha calculation page:
-13-26i - Wolfram|Alpha
it factorises -13-26i as
i×(1+2 i)×(2+3 i)×(3+2 i)
my question is why is the i a necessary factor to take out? are the rules regarding this?
Thanks
I don't think that this is correct. There is a characterisation of Gaussian primes that says that a Gaussian integer is prime if and only if one of the conditions is true:
- Its norm is prime (as an integer)
- It is a prime (as an integer) of the form $\displaystyle 4k+3$
In this case, the norm of $\displaystyle 1-i$ is $\displaystyle 1^2+(-1)^2=2$, which is prime. So $\displaystyle 1-i$ is prime.
The issue that should be brought up is that unique factorisation only holds up to a unit (i.e. invertible number). For example, normal integer factorisation follows this idea: $\displaystyle 25=5\times 5$ or $\displaystyle 25=(-5)\times (-5)$. In the second example, we have multiplied the factors by $\displaystyle -1$ (which is invertible), but we do not consider it to be inherently different from the first. So in this case, $\displaystyle 1-i$ only differs from $\displaystyle 1+i$ by multiplying by the unit $\displaystyle i$, which is no difference at all, as far as unique factorisation goes.
So to address original question, throwing in $\displaystyle i$ is analogous to repeatedly multiplying by $\displaystyle -1$ with the usual integer factorisation. If it bothers you, you may let one of the terms absorb it.