How does this factoring work?

Question

$$ (z^2 - 2i ) = (z -1 -i)(z + 1 +i) $$

I see if you multiply out the right-hand side, you obtain the left-hand side, but how does one know to factor like that or this?

$$ (z^2 − 3iz − 3 + i) = (z − 1 − i)(z + 1 − 2i) $$

Answer 1

The first one is using $z^2-a^2 =(z-a)(z+a) $ where $a = \sqrt{2i} =1+i $ since $(1+i)^2 =1+2i-1 = 2i $.

The second one just uses the quadratic formula, which works for complex as well as real coefficients, to solve $z^2 − 3iz − 3 + i = 0 $. If the roots are $u$ and $v$, then $z^2 − 3iz − 3 + i = (z-u)(z-v) $.

(This was added later)

Using the quadratic formula, the roots are

$\begin{array}\\ \dfrac{3i\pm\sqrt{(-3i)^2-4(-3+i)}}{2} &=\dfrac{3i\pm\sqrt{-9+12-4i}}{2}\\ &=\dfrac{3i\pm\sqrt{3-4i}}{2}\\ &=\dfrac{3i\pm(2-i)}{2} \qquad\text{since }\sqrt{3-4i} = 2-i\\ &=\dfrac{3i+(2-i), 3i-(2-i)}{2}\\ &=\dfrac{2+2i, -2+4i}{2}\\ &=1+i, -1+2i\\ \end{array} $

Answer 2

If we are trying to factor $p(z) = z^2 - 2i$ we can determine the factorization by finding the zeros of the polynomial.

If $p(z) = 0$ then $z^2 = 2i$. Thus $|z| = \sqrt{2}$. Writing $z=\sqrt{2} e^{i\theta}$ we have $e^{i2\theta} = e^{i \frac{\pi}{2}}$. This yields $$\cos(2\theta) + i \sin(2\theta) = \cos(\pi/2) + i \sin(\pi/2).$$

Now if we restrict $\theta \in [-\pi,\pi]$, we find the solutions $\theta = \pi/4$ and $\theta = -\frac{3\pi}{4}$.

This means $p(\sqrt{2} e^{i \pi/4} ) = 0$ and $p(\sqrt{2} e^{-i\frac{3\pi}{4}})$, and since it is monic, it can be factored as $$(z- \sqrt{2} e^{i\pi/4}) ( z - \sqrt{2} e^{-i3\pi/4}).$$

Finally $\sqrt{2} e^{i \pi/4} = \sqrt{2} ( \cos(\pi/4) + i \sin(\pi/4) ) = 1 + i$, and $\sqrt{2} e^{i3\pi/4} = -1-i$.