TITLE: How to solve $z^4 +6z^2 +25 = 0$ into complex conjugate?
QUESTION [5 upvotes]: How to solve $z^4 +6z^2 +25 =  0$ into complex conjugate?
I started with
$$(z^2 + 3)^2 + 16 = 0$$
$$(z^2 + 3)^2  = - 16$$
$$z^2 + 3 = \pm 4i$$
Is this the way to start solving the equation or am I completely of?

REPLY [3 votes]: The aim is to find a good representation of the square roots of $-3 \pm 4i$. But these are Gaussian integers, members of a unique factorization domain, so you should try to factor them into primes. Voilà: $-3+4i = (1+2i)^2$ and $-3-4i=(1-2i)^2$. So the solutions of your equation are $\pm1\pm2i$.
The method? The Gaussian primes are $1+i$ (dividing $2$), the ordinary primes congruent to $3$ modulo $4$, and for each prime $p\equiv 1\bmod 4$, a pair of primes $\pi_1$ and $\pi_2 = \overline{\pi_1}$ with $\pi_1\pi_2=p$. This splitting of $p$ arises out of the theorem that such $p$ is the sum of two squares. In particular, $5=(1+2i)(1-2i)=1+4$. So you look at $-3+4i$ and notice that $(-3+4i)(-3-4i)=25=(1+2i)^2(1-2i)^2$ and use Unique Factorization, but judiciously, 'cause you have to take into account the fact that there are four units, $\pm1$ and $\pm i$.