Confusion about roots

Question

Solving $$z^4 + 1 = 0$$ where $z = x+iy$, I tried by factorizing the expressiong into $$z^4 + 1 = (z^2 - 2z + 2)(z^2 + 2z + 2).$$ But then solving $$z^2 - 2z + 2 = 0$$ for $z$ gives the solutions $z = 1 \pm i$, and solving $$z^2 + 2z + 2 = 0$$ for $z$ gives the solutions $z = -1 \pm i$. The actual roots should be $z = \frac{1 \pm i}{\sqrt{2}} $ and $ z= \frac{-1 \pm i}{\sqrt{2}}$. I'm possibly missing something that should be obvious to me here, but what about $1/\sqrt{2}$?

Answer 1

Your starting factorization is not correct. It should be:

$$z^4+1 = z^4+2z^2+1-2z^2 = (z^2+1)^2-(\sqrt{2}z)^2=...$$

Answer 2

Are you sure about that factoring? Have you multiplied out $(z^2 - 2z + 2)(z^2 + 2z + 2)$ and seen what you get? Just the constant term should be enough to give it away, really.

The simplest solution in my opinion is to rewrite the equation to $$ z^4 = -1 $$ and then think geometrically in the complex plane from there. If you really want a more algebraic solution, then I would first note that $z^4 = (z^2)^2$, and get $$ z^4+1 = (z^2)^2+1 = (z^2+i)(z^2-i) $$ which to me is a mich nicer factorisation into quadratics, even though complex numbers appear.