<s>[INST] Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as  $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadratic formula. However, I'm not sure if there exists any method to factor $z^2 - 2i$ and $z^2 + 2i$? 
I would greatly appreciate it if people could please explain how one would go about this.
 [/INST] You can find roots of $z^4+4=0$. For example, one easy way is to multiply the equation by $z^4-4$ to get $z^8 - 16 = 0$, so we conclude that roots of $z^4+4$ are $8$th roots of unity (times $\sqrt 2$) that are not $4$th roots of unity, i.e. the roots are $$\{\omega\sqrt 2,\omega^3\sqrt 2,\omega^5\sqrt 2,\omega^7\sqrt 2\}$$ where $\omega = e^{i\pi/4} = \frac{\sqrt 2} 2+i\frac{\sqrt 2} 2.$ Multiplying by $\sqrt 2$ and rotating in complex plane we get that the roots are $$\{1+i,-1+i,-1-i,1-i\}$$ and finally $$z^4+4=(z-1-i)(z+1-i)(z+1+i)(z-1+i).$$
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