Why is purely imaginary exponential growth at a constant rate in the orthogonal direction instead of an accelerating rate of circular motion?

Question

Eulers identity gives us a manner to describe traversing a circle using imaginary numbers $e^{ix}=\cos(x)+i\cdot \sin(x)$.

From what I understand, the right part is rather simple and just describes a location on a circle using a complex number. The left part though apparently traverses us around the circle at constant increments each constantly changing the direction by 90 degrees. If you think about tracing out the unit circle in small increments using the left hand side of the equation, from what I understand the transformation to each subsequent increment will necessarily be an operation which brings you to a position orthogonal to the starting point because its a circle. However this rate at which the tracing occurs shouldn't speed up. I think I understand this mental picture but don't understand the intuition that we don't have an acceleration for the rate we traverse the circle. It is exponential after all. I am looking for an intuitive approach to understanding the left side.

Answer 1

There's an elegant connection to the polar representation of motion in a plane. Associate $x+iy$ with $x\vec{i}+y\vec{j}$ so $re^{i\theta}$ becomes $r\hat{\vec{r}}$ and $ie^{i\theta}=\hat{\vec{\theta}}$. With $\dot{\theta}=1$, the velocity is then the purely transverse vector $ir\hat{\vec{r}}=r\hat{\vec{\theta}}$. Why is it orthogonal to the radius? Because vectors $\vec{w},\,\vec{z}$ associated with complex numbers $w,\,z$ satisfy $\vec{w}\cdot\vec{z}=\Re (\overline{W} z)$. And of course, $\Re (\overline{e^{i\theta}} ie^{i\theta})=\Re i=0$.

Answer 2

There is a difference between speed and velocity. Speed is a scalar and velocity is a vector quantity. Thus, circular motion at constant speed implies velocity which is perpendicular to position vector from the origin and which changes direction and hence centripetal acceleration or force in a perpendicular direction to the motion towards the center is required to keep it moving on the circle.

The mathematical interpretation of this is very similar. That is, if $\;p(x) := e^{ix}\;$ then $\;p'(x) = i\;e^{ix}.\;$ Multiplication by $\;i\;$ is just rotation by $90$ degrees so the velocity vector is always perpendicular to the position vector. This implies that the magnitude of the position vector remains constant. A very similar agrument about the velocity vector shows its magnitude remains constant also.