What is the meaning of having a position vector $\vec r = z_1 \hat i +z_2 \hat j +z_3 \hat k$ where $z_1,z_2$ and $z_3$ are complex numbers?

Question

What is the geometrical interpretation of this?

I came across this as a solution to a second order differential equation when I was solving a problem including $m\vec a = f(\vec r)$.

Thank you for taking out time to read this.

Answer 1

Let $\vec \alpha$ be a non-zero vector in a plane.

$\iff \exists$ a unique point $P$ on the plane such that $\vec \alpha=\vec {OP}$ (position vector)

$\iff \exists$ a unique ordered pair $(x,y)$ $(\neq (0,0))$ of real numbers $x,y$ such that $\vec \alpha=x\hat i+y\hat j$

Again, given a point $P$ in the plane distinct from $O$, $P$ has the cartesian coordinates $(x,y)$

$\iff P$ is represented by the complex number $x+iy$

$\iff P$ has polar coordinates $(r,\theta)$, where $x=r\cos \theta,y=r\sin \theta$ with $r>0$, $-\pi<\theta \leq \pi$

$\iff P$ has position vector $x\hat i+y\hat j$

This unifies the concepts of vectors with those of complex numbers, cartesian coordinates and polar coordinates. In your case, we can extend this to three dimensions as well.

Answer 2

It makes sense to have a three dimensional vector space over the complex numbers. That would be equivalent to a 6 dimensional vector space over the real numbers.

Answer 3

In response to your comment, the problem you were considering is a two dimensional harmonic oscillator in three dimensions. It's sometimes called a harmonic potential.

Given the potential $V=x^2+y^2$, Newtons second law tells us to solve

$$m\vec{a}=m\ddot{\vec{r}}=-\nabla V(\vec{r})$$

for the eom with some initial conditions. With $f(\vec{r})=-\nabla V=(-2x,-2y,0)$ this decouples the differential equation into three odes.

$$\begin{align} m\ddot{x}&=-2x\\ m\ddot{y}&=-2y\\ \ddot{z}&=0\\ \end{align}$$

The solution of the first two equations will involve sines and cosines, and what is frequently done when dealing with waves, vibrations, electromagnetic waves etc. is to write the equations in terms of complex numbers:

$$\begin{align} m\ddot{z_1}&=-2z_1 \quad \text{where } x=\Re(z_1)\\ m\ddot{z_2}&=-2z_2 \quad \text{where } y=\Re(z_2)\\ \ddot{z_3}&=0 \quad \text{where } z=\Re(z_3)\\ \end{align}$$

The idea is that some manipulations like adding or superimposing vibrations is easier to manipulate when using complex numbers using the identity $e^{i\theta}=\cos \theta +i\sin\theta$. After we do our manipulations then we just take the real part of the equations again.

Solving the complex equations:

$$\begin{align} z_1&=A_1 e^{it\sqrt{2/m}}+ A_2 e^{-it\sqrt{2/m}} \\ z_2&=B_1 e^{it\sqrt{2/m}}+ B_2 e^{-it\sqrt{2/m}}\\ z_3&=C_1t+C_2 \end{align}$$ where the letters are constant complex numbers.

You could collect the $z$'s into a vector whose real part is the original vector, $\vec{w}(t)=(z_1,z_2,z_3)$.

We can recover the solutions to the original differential equation via:

$$\begin{align} x(t)&=\Re(A_1+A_2)\cos(t\sqrt{2/m})-\Im(A_1-A_2)\sin(t\sqrt{2/m})\\ y(t)&=\Re(B_1+B_2)\cos(t\sqrt{2/m})-\Im(B_1-B_2)\sin(t\sqrt{2/m})\\ z(t)&=\Re(C_1)t+\Re(C_2) \end{align}$$

It's probably a bit unnecessary for this example $\ldots$

In the case for example when $A_1=ae^{i\alpha}$, $A_2=ae^{-i\alpha}$ we would have $$z_1(t)=2a\cos(t\sqrt{2/m}+\alpha)=x(t)$$ and so on.

French's book 'Vibrations and Waves' has a good few examples as to why you might want to work with complex numbers, and in a tex like Griffiths 'Introduction to Electrodynamics' discusses writing the $E$ and $B$ fields as complex fields whose real parts are the $\textit{physical}$ electromagnetic fields.

Summary It's mostly done for convenience in calculation.