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.
 A: 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.
A: 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.
