Consider a set of complex numbers such that $z=a+ib$ such that $a$ and $ b$ is an integer.
Naturally the product of two set elements is another element of the set. Now, let us try to determine the factorization of $z_3$ using:
$$z_1 z_2 = z_3$$
Method $1$
Take the modulus
$$\implies |z_3|^2 = |z_1|^2 |z_2|^2 $$
Using prime factorization of $|z_3|^2$ we can guess the prime factors of $|z_1|^2$ and $|z_2|^2$ and hence, we will have many cases of $|z_1|^2$ and $|z_2|^2$ depending on the number of prime factors.
Then one (or more) of the cases will be resolvable into $|z_i|^2 = \Re(z_i)^2 + \Im(z_i)^2$ such that the real $ z_{i.R} $ and imaginary part $ z_{i,I} $ of $z_i$ is an integer (where $i = 1$ or $2$) out of those solutions we will see which ones satisfy $z_1 z_2 =z_3$
Example $1$
Let $z_3 = -6 + 17i \implies |z_3|^2 = 325 = 5^2 \times 13$
We can decompose this into $ |z_1|^2 = 5 $ and $ |z_2|^2 = 5 \times 13 $ or $ |z_1|^2 = 25 $ and $ |z_2|^2 = 13 $
We note $5^2 = 3^2 + 4^2$ and $13 = 2^2 + 3^2$
Hence, using hit and trail the numbers are: $ z_1 = 2+3i$ and $z_2 = 3+4i$
Method $2$
We consider:
$$ z_1 = a +ib$$
$$ z_2 = c + id$$
$$ z_3 = e + if$$
$$\implies |z_1|^2(c+id) = (e+if)(a-ib)$$
Comparing real and imaginary parts:
$$ |z_1|^2 c + i |z_1|^2 d = (ae +fb) + i (fa – be)$$
Comparing real and imaginary parts:
$$|z_1|^2 =\frac{(ae+fb)}{c} = \frac{fa – be}{d}$$
$$\implies cd (a^2 +b^2)^2 = (ae + fb)(fa – be) $$
Hence, we can guess again using prime factorization.
Questions
- Are there other useful methods of factorizing complex numbers?
- Are these methods known(/correct/useful)?
- Am I being funny or is there a hidden(deeper) statement about complex prime factorization statement that can be made?
- Hence we have $2$ methods to factorize $z_3$
Essentially what the picture above shows us is that the using method 1 one can convert the complex $z_3$ factorization problem into a sum of squares and a real prime factorization problem(s). However using method $2$ it is only a real prime factorization problem(s). Is it possible to somehow use this to convert a sum of squares problem as a prime factorization problem?
Best Answer
Lets think about Gaussian integer : $\Bbb{Z}[i]=\{a+ib : a,b\in\Bbb{Z}\}.$
This is a ring under usual complex number addition and multiplication and here prime numbers are (up to multiplication by $i$ )
For example: $90$ has the prime factorization $3^2(1+i)(1-i)(2+i)(2-i).$ We can use this factorization to find all the representations $90$ as a sum of two squares as $$90=3^2(1^2+1^2)(1^2+2^2).$$
To find the prime factorization in $z\in\Bbb{Z}[i]$ :
First find $|z|.$ If this is a prime in $\Bbb{Z}$ of the form $4n+3,$ then $z$ is already a prime. Otherwise look at the prime factorization of $|z|$ in $\Bbb{Z}.$
For example: Consider $1+8i$ which has $|1+8i|=5\times 13$ and $5=2^2+1^2=(2+i)(2-i)$ and $13=2^2+3^2=(2+3i)(2-3i)$ now using those factorizations, we can find out that $$1+8i=(2+i)(2+3i).$$