I have a complex polynomial $z^2 + 2z + 2$.
From trying out roots repeatedly I can see the correct roots are
$z = -1 + i$ and
$z = -1 – i$
But what is the standard method for getting roots of a complex polynomial like this?
[Math] Roots of complex polynomial
algebra-precalculuspolynomials
Related Solutions
The matrix method does not simplify the matter at all (unless you have a mental block about polynomials but feel really comfortable about matrices; or unless you have some really cool method for finding eigenvalues that does not depend on the characteristic polynomial).
Not only did you still get a polynomial of degree $4$, you got essentially the same polynomial you started with! To see this, let $p(t)$ be your original polynomial, and let $q(\lambda)$ be the characteristic polynomial you found. Then you have: \begin{align*} -3q(\lambda) &= -3\lambda^4 - 5\lambda^3 -10\lambda^2 -20\lambda + 8\\ &= \lambda^4\left(-3 -5\left(\frac{1}{\lambda}\right) - 10\left(\frac{1}{\lambda^2}\right) - 20\left(\frac{1}{\lambda^3}\right) + 8\left(\frac{1}{\lambda^4}\right)\right)\\ &= \lambda^4p\left(\frac{1}{\lambda}\right) \end{align*}
That means that if $r$ is a root of $p(t)$, then $\frac{1}{r}$ is a root of $q(\lambda)$ (note that $r=0$ is not a root) and conversely; so finding the roots of the characteristic polynomial of the matrix is the same as finding the roots of your original $p(t)$.
This will always happen with this method: the matrix you write is essentially just the companion matrix of the original polynomial, so the characteristic polynomial will essentially be the original polynomial again (or perhaps a reversal of the powers, as above). But for any polynomial $p(x) = a_nx^n + \cdots +a_0$ with $a_na_0\neq 0$, the roots of $p(x)$ and the roots of $x^np(\frac{1}{x}) = a_n + a_{n-1}x + \cdots + a_0x^n$ are just reciprocals of each other, so finding roots for the former is essentially the same as finding roots for the latter.
Added: As Andres mentions and Qiaochu hints, it's possible that the change of perspective will let you more easily apply certain algorithms or theorems by going to the matrix rather than dealing with the polynomial directly. But my point is that the matrix method only affords you a change of perspective rather than a simplification or an essential change in the problem. You are still dealing with the same polynomial when everything is said and done.
The rational root method is pretty good for polynomials with relatively small integer coefficients; for polynomials of degree $3$ or $4$ you can always try the formulas by radicals (though they can be pretty annoying to actually use). There are other algorithms for finding, or at least approximating, roots of polynomials; the link Qiaochu gives in the comments is a good place to start.
The conjugate factor theorem states that for a polynomial $p(x)$ with real coefficients, the complex roots come in conjugate pairs, or if $a+bi$ is a root, then $a-bi$ is also a root. From this we see that your polynomial has the roots $$\frac{2}{3}, -1, 3+\sqrt{2}i, 3-\sqrt{2}i.$$ Therefore, $$(x-\frac{2}{3})(x+1)(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0.$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x^2-6x+11)=0.$$ Continuing the expansion results in the polynomial $$p(x)=\frac{1}{3}(3x^4-17x^3+25x^2+23x-22).$$
Best Answer
The quadratic formula works! You just need to interpret the square root as the complex square root. So the two roots are given by the two values of $$\frac{-2 +\sqrt{4-8}}{2} = -1 +\frac{\sqrt{-4}}{2} = -1+\sqrt{-1}.$$ Since $\sqrt{-1}$ takes two values, $i$ and $-i$, the two roots are $-1-i$ and $-1+i$.
Complex roots to polynomials with real coefficients always come in "conjugate pairs"; that is, if $a+bi$ is a root, so is $a-bi$.
The same "tricks" as for real numbers also work for complex numbers: if $r_1$ and $r_2$ are the roots of $az^2+bz+c$, then we will have $az^2+bz+c = a(z-r_1)(z-r_2)$. So you are looking for numbers $r_1$ and $r_2$ such that $r_1r_2 = \frac{c}{a}$ and $r_1+r_2 = -\frac{b}{a}$. Here, you are looking for two complex numbers whose product is $2$ and whose sum is $-2$. If you can find them "by inspection" (staring at the polynomr ealizing what the right answer is), then that's it.Since the polynomial has real coefficients, either you have two real roots which it is easy to verify don't exist), or they are two complex numbers of the form $a+bi$ and $a-bi$, with $a,b$ reals. So we need $(a+bi)+(a-bi) = -2$ and $(a+bi)(a-bi) = 2$. Or, $2a=-2$, $a^2+b^2 = 2$. Hence $a=-1$, $b^2=1$, giving the two solutions.