We will see that the 3D plot of $$x^2 + (y+zi)^2 = 1$$ contains a circle and a hyperbola, where $$i$$ is the imaginary number. Beyond the visuals, this helps us understand the (complex) eigenvalues of real matrices.

If we expand out $$x^2+(y+zi)^2 = 1$$ in real and imaginary parts, we get

\begin{align*} &\text{Real Part:} &x^2 + y^2 - z^2 &= 1, \\ &\text{Imaginary Part:} &yz &= 0. \end{align*}

From $$yz=0$$ we get two cases.

\begin{align*} \text{Case 1: }y&=0 \text{ and so } x^2-z^2 = 1\\ \text{Case 2: }z&=0 \text{ and so } x^2+y^2 = 1 \end{align*}

Case 1 nets us a hyperbola in the $$xz$$-plane. Case 2 nets us a unit circle in the $$xy$$-plane.

You can check out the plot at desmos.

Eigenvalues and dynamical systems

Why might we care about this beyond the visuals? This sort of graph often comes up when we’re trying to understand the (complex) eigenvalues of a real matrix that depends on a real parameter. For example, consider the matrix

$M(\mu) = \begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}$

for some real parameter $$\mu$$. The eigenvalues $$\lambda$$ of $$M$$ satisify the equation $$\mu^2 + \lambda^2 = 1.$$ Letting $$\mu = x$$ and $$\lambda = y+zi$$ recovers the plot of $$x^2 + (y+iz)^2 = 1.$$

Just from looking at the plot, we see that

\begin{align*} \mu < -1&: M(\mu) \text{ has a pair of complex conjugate eigenvalues} \\ \mu = -1&: M(\mu) \text{ only has one eigenvalue}, \lambda = 0 \\ 0< \mu < 1&: M(\mu) \text{ has one +ve and one -ve eigenvalue} \\ \mu = 1&: M(\mu) \text{ only has one eigenvalue}, \lambda = 0 \\ \mu > 1&: M(\mu) \text{ has a pair of complex conjugate eigenvalues} \end{align*}

We are sometimes interested in the magnitude of the eigenvalues. We can see that $$\vert \lambda \vert = \sqrt{y^2+z^2}$$ is the radius of the $$yz$$-plane. Again just from the plot we can guess / see that

\begin{align*} \vert\lambda\vert\leq1 &\iff \mu \in [-\sqrt{2}, \sqrt{2}], \\ \vert\lambda\vert<1 &\iff \mu \in (-\sqrt{2}, \sqrt{2}) \backslash \{0\}. \end{align*}

Understanding of the eigenvalues of $$M(\mu)$$ is generally useful if $$M(\mu)$$ shows up in a dynamical system. For example let’s consider the system $$\mathbf{v}_{n+1} = M(\mu)^2 \mathbf{v}_n$$, then

\begin{align*} \mathbf{v}_{n+1} &= M(\mu)^2\begin{bmatrix} x \\ y \end{bmatrix} \\ &=\begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}\begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} \\ &=\begin{bmatrix} 0 & 1+\mu \\ 1-\mu & 0 \end{bmatrix}\begin{bmatrix} (1+\mu)y \\ (1-\mu)x \end{bmatrix}\\ &= (1-\mu^2) \begin{bmatrix} x \\ y \end{bmatrix} \\ &= (1-\mu^2)\mathbf{v}_n. \end{align*}

Curiously, we see that the origin is a stable fixed point if and only if $$\vert1-\mu^2\vert \leq 1$$, which is equivalent to $$\mu \in [-2,2]$$ as well as $$\vert\lambda\vert \leq 1$$. Similarly, the origin is an asymptotically stable fixed point if and only if $$\vert 1-\mu^2\vert < 1$$ if and only if $$\vert \lambda \vert < 1.$$ This hints at a relationship between eigenvalues and stability.

More examples

Consider the matrix

$M(\mu) = \begin{bmatrix} 1 & \mu \\ 1 & 1 \end{bmatrix}.$

We can notice a couple of things immediately. When $$\mu=1$$, $$M(\mu)$$ is degenerate and so has a zero eigenvalue. When $$\mu=0$$, $$M(\mu)$$ is a shear matrix with only one eigenvalue $$\lambda = 1$$. As shear matrices live on the boundary of diagonalisable and non-diagonalisable matrices, we can reasonably further guess that if $$\mu<0$$ then $$M(\mu)$$ has complex conjugate eigenvalues and if $$\mu>0$$ then $$(M\mu)$$ has real eigenvalues.

Concretely, considering the eigenvalue equation $$\lambda^2 - 2\lambda + (1-\mu) = 0$$ and letting $$\mu=x$$ and $$\lambda=y+zi$$ and re-arranging terms we get

$\left((y+zi)-1\right)^2=x.$

By splitting into real parts and imaginary parts and re-arranging terms we get

\begin{align*} &\text{Real Part:} &(y-1)^2 - z^2 &= x, \\ &\text{Imaginary Part:} &(y-1)z &= 0. \end{align*}

Again we have two cases

\begin{align*} \text{Case 1: }y&=1 \text{ and so } -z^2 = x\\ \text{Case 2: }z&=0 \text{ and so } (y-1)^2 = 0 \end{align*}

Case 1 nets us a parabola in the $$xz$$-plane while Case 2 nets us a parabola in the $$xy$$-plane. Finally, we arrive at the plot.

You can create these sort of plots for any other 2 by 2 matrix that depends on one real parameter using a desmos calculator I made.