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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.

Joe
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    “The left part though apparently traverses us around the circle at constant increments each constantly changing the direction by 90 degrees.” No idea what you mean by this. – spaceisdarkgreen May 15 '18 at 19:07
  • I think based on your following text, you might mean “rotates around the unit circle at constant speed” but that seems like a manifest property of the right hand side, not the left. (Though of course since the left is equal to the right, it holds for the left too.) – spaceisdarkgreen May 15 '18 at 19:09
  • 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 – Joe May 15 '18 at 19:13
  • A physicist would say that as the acceleration is perpendicular to the velocity, no work is done, and so the kinetic energy remains constant. – Angina Seng May 15 '18 at 19:28
  • I think my use of acceleration threw us off into a physical explanation where I'm looking for a mathematical one. I am not trying to get an explanation on uniform circular motion... I am fine with that from my physics classes. I understand speed velocity acceleration etc. I would like to direct the answers back toward the initial imaginary exponential growth and why we don't get faster rotation – Joe May 15 '18 at 19:31

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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.

Somos
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  • I think my use of acceleration threw us off into a physical explanation where I'm looking for a mathematical one. I am not trying to get an explanation on uniform circular motion... I am fine with that from my physics classes. I understand speed velocity acceleration etc. I would like to direct the answers back toward the initial imaginary exponential growth and why we don't get faster rotation – Joe May 15 '18 at 19:32
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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, $\Re (\overline{e^{i\theta}} ie^{i\theta})=\Re i=0$.

J.G.
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