Is there a relationship between rotations in $\mathbb{R}^2$ and polynomial roots?

Question

Given that a complex number $z = a + bi$ can be described as a matrix:

$$\begin{bmatrix}a & -b\\b & a\end{bmatrix}$$

and, through Euler's formula, if $|z| = 1$ this matrix becomes the rotation matrix:

$$\begin{bmatrix}\cos(\theta) & -\sin(\theta)\\\sin(\theta) & \cos(\theta)\end{bmatrix}$$

it's clear that complex numbers represent rotations in $\mathbb{R}^2$. Complex numbers are also roots of polynomial equations with complex coefficients. I was just wondering if there is any deeper relationship between the two concepts, or if it's just a convenient happenstance that rotations are represented through complex numbers.

Answer 1

Interesting question. First, let's straighten some stuff out.

Multiplication by a complex number represents a rotation and a rescaling. This is because when you multipling two complex numbers is the same as multiplying their magnitudes and adding their angles. This follows directly from the polar form of complex numbers:

$$ (r_1 e^{i \theta_1})(r_1 e^{i \theta_1}) = r_1 r_2 e^{i (\theta_1+\theta_1)} = r_3 e^{i \theta_3} $$

In Cartesian coordinates the multiplication looks like this:

$$ (a_1 + i b_1 )(a_2 + i b_2 ) = (a_1 a_2 - b_1 b_2 ) + i ( a_1 b_2 + a_2 b_1 ) = a_3 + i b_3 $$

Now, if you treat the complex value on the complex plane as a vector instead, the equivalent matrix operation is:

$$ \left[\begin{array}{c} a_3 \\ b_3 \end{array}\right] = \left[\begin{array}{c} a_1 a_2 - b_1 b_2 \\ a_1 b_2 + a_2 b_1 \end{array}\right] = \left[\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right] \cdot \left[\begin{array}{c} a_2 \\ b_2 \end{array}\right] $$

So that's where your "complex number described as matrix" comes from and is applicable in the context of multiplication.

Polynomials with real coeffictients can have complex roots as well, so there is a slight misstatement in your question. If $r$ is a root of $P(x)$, that means that $x-r$ is a factor of $P(x)$, so when $x=r$ then $P(x)=0$.

Since $x-r$ is a subtraction and not a multiplication, I don't really see a relationship between the rotation representations of the roots and the polynomial itself. You are also ignoring the rescaling in this consideration.

Hope this helps.

Ced