Is a complex number treatment of 2D vectors valid?

For example, can I treat a 2D position vector as a complex number instead of a vector while trying to derive the formula for centripetal acceleration in uniform circular motion.

$$\mathbf{r}=re^{i\theta}\\ \mathbf{\dot{r}}=rie^{i\theta}\dot{\theta}\\ \mathbf{\ddot{r}}=-re^{i\theta}\dot{\theta}^2$$

Which is an acceleration directed anti-parallel to the direction of the position vector( i.e. towards center) and of the magnitude of $r\dot{\theta}^2$

I did this because it's easier to differentiate $e^{kx}$ instead of keeping track of the signs of the sines and the cosines.

• Yes, as long as you stay in two dimensions this is perfectly valid. In more complicated cases it might not be so easy to separate tangential and normal components, though. – Javier Apr 3 '17 at 18:33
• Comment #1: your equations above assume that $r$ is a constant. This is fine, but it should be stated explicitly, and might not be true in some circumstances. – Michael Seifert Apr 3 '17 at 19:50
• Comment #2: this sort of trick is used all the time in physics. It becomes particularly nice when dealing with 2D problems in which magnetic fields or Coriolis forces are important, and you need to take cross products. It works out that the cross product of any vector in the $xy$-plane with a vector in the $z$-direction is equivalent to just multiplying the corresponding complex number by $i$. – Michael Seifert Apr 3 '17 at 19:54
• I assumed $r$ as a constant because I was deriving this for Uniform Circular Motion. – Archimedesprinciple Apr 4 '17 at 4:17

Differentiation doesn't involve imaginary unit:

$$\left.\frac{df}{dx}\right|_{x_0}\triangleq\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}.$$

Moreover, it's also linear, i.e. $$\frac{d(af(x)+bg(x))}{dx}=a\frac{df(x)}{dx}+b\frac{dg(x)}{dx}.$$

This means that both real and imaginary part of the function are differentiated independently. Thus, if your motion in $x$ and $y$ coordinates is represented by a function like $r=x+iy$, then the derivative of this function will be $\dot r=\dot x+i\dot y$, which is exactly what you want.

So yes, such treatment is valid.

I think this s an awful approach (and as I explain below, the wrong solution to your real issue).

In your definition the problem is that :

$$\mathbf r \cdot \mathbf r \neq r^2$$

Which to be a valid representation, it should equal.

You therefore need special care to handle your representation.

I did this because it's easier to differentiate $e^{kx}$ instead of keeping track of the signs of the sines and the cosines.

Honestly this is a very poor reason. You need to develop the (not very difficult) skill of becoming familiar with the trigonometric functions and manipulating them, not avoiding them.

Signs are just too important (especially in physics) to spend your time avoiding.

• I'd disagree that this is an awful approach. It really simplifies some calculations. In fact, even Landau in his books does use this in a couple of places (e.g. motion of a charge in constant magnetic field — $\S21$ in The Classical Theory of Fields), and I wouldn't suppose he wasn't familiar with trigonometric functions. – Ruslan Apr 3 '17 at 19:34
• @StephenG I didn't take any dot products. I differentiated. I was treating $r$ as a constant because this was for circular motion. – Archimedesprinciple Apr 4 '17 at 4:22
• @StephenG Also, I tried the same thing for varying $r$ and $\theta$ – Archimedesprinciple Apr 4 '17 at 5:05