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.

  • $\begingroup$ The question is vague enough it's not clear whether this is an answer, but you might note that the eigenvalues of that matrix are $e^{\pm i\theta}$. $\endgroup$ – David C. Ullrich May 4 '18 at 15:47

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.


  • $\begingroup$ Interesting answer. But I might argue that a root of a polynomial that has magnitude 1 in the complex plane represents a rotation. That's sort of what I was getting at: is it just that the roots are complex numbers, and complex numbers also represent rotations, or is there some theory of polynomials that relates certain classes of polynomials to sets of rotations that are interesting? $\endgroup$ – Michael Stachowsky May 4 '18 at 16:33
  • $\begingroup$ @MichaelStachowsky, I am not aware of any, but that doesn't mean much. Of course the Roots of Unity ($z^N-1=0$) are a set of N evenly spaced points on he complex unit circle, so each has magnitude 1. Maybe you'll find something pertinent to you in here: en.wikipedia.org/wiki/Properties_of_polynomial_roots $\endgroup$ – Cedron Dawg May 4 '18 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.