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.
On
Using difference of squares ...
$$z^2-2i= z^2 - 2e^{i \frac \pi 2} = z^2 - (\sqrt 2 e^{i \frac \pi 4})^2 \\ = (z+\sqrt 2 e^{i \frac \pi 4})(z-\sqrt 2 e^{i \frac \pi 4})$$
$$z^2+2i= z^2 - 2e^{i \frac {3\pi} 2} = z^2 - (\sqrt 2 e^{i \frac {3\pi} 4})^2 \\ = (z+\sqrt 2 e^{i \frac {3\pi} 4})(z-\sqrt 2 e^{i \frac {3\pi}4})$$
On
For example: $$z^2-2i=z^2-(1+i)^2=(z-1-i)(z+1+i).$$ I think the following a bit of better. $$z^4+4=z^4+4z^2+4-4z^2=(z^2+2)^2-(2z)^2=$$ $$=(z^2-2z+2)(z^2+2z+2)=((z-1)^2-i^2)((z+1)^2-i^2)=$$ $$=(z-1-i)(z-1+i)(z+1-i)(z+1+i).$$
On
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 $$z^4+4=(z-1-i)(z+1-i)(z+1+i)(z-1+i).$$
Solving $$z^2-2i=0$$
we have $$z^2=2i=2\exp\left( \frac{i\pi}{2}\right)$$
$$z=\pm\sqrt2\exp\left( \frac{i\pi}{4}\right)=\pm\sqrt{2}\left(\cos\left( \frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4} \right)\right)=\pm(1+i)$$
Hence, $$z^2-2i=(z-(1+i))(z+(1+i))$$
Try the same trick on $z^2+2i$.