$e^{i \pi} = -1$. I get why this works from a sum-of-series perspective and from an integration perspective, as in I can evaluate the integrals and find this result. However, I don't understand it intuitively.
Intuitively, what this means to me is that if you rotate pi radians around a unit circle, you will end up exactly opposite of where you started.
Expanding upon this, for any theta, $e^{i\theta}$ is equivalent to rotating $\theta$ radians around a unit circle. So it's obvious (if not intuitive) to me that $e^{(\pi/2)i} = i$ and $e^{2\pi i} = 1$ and so on.
What I'm wondering is, intuitively, how is the natural logarithm related so closely to circles? My understanding of $e$ stems from exponential growth, and I don't see how that ties to rotation around a unit circle.
I know the formulas, but I'm looking for an intuitive explanation. For example, when I used to ask how sin and cos related to circles, people would show me taylor series or tables with a bunch of values or tell me to use a calculator, but it didn't click until somebody told me that sin is just a measure of the height of a point as you travel around the unit circle. Then it all makes sense.
I'm looking for that kind of explanation of $e$ - how is $e$ related to circles and why does $e^{ix} = \cos(x) + i\sin(x)$?
Answer
There is a closely related discussion at this math.SE question with lots of details. Let me see if I can summarize it as concisely as I can:
- Let $r(t) : \mathbb{R} \to \mathbb{R}^2$ parameterize a particle moving uniformly around the unit circle. Then $|r(t)|^2 = \langle r(t), r(t) \rangle = 1$ (where $\langle \cdot, \cdot \rangle$ denotes the dot product). Differentiating this relation gives $\langle r(t), r'(t) \rangle = 0$; in other words, the displacement is always orthogonal to the velocity. (This should make physical sense.)
- Since in addition $r(t)$ is moving uniformly we have $|r'(t)|^2$ is a constant, and we may as well assume $|r'(t)| = 1$ by a suitable change of units. Hence $r'(t)$ is either a $90^{\circ}$ clockwise or counterclockwise rotation from $r(t)$.
- Now identify $\mathbb{R}^2$ with $\mathbb{C}$ and identify $r$ with a function $z(t) : \mathbb{R} \to \mathbb{C}^2$. If the particle is moving counterclockwise, then the above implies that $z'(t) = i z(t)$.
- But this differential equation clearly has unique solution $z(t) = e^{it} z(0)$.
So the fact that multiplication by $i$ is the same as rotation by $90^{\circ}$ actually immediately implies the more general fact about arbitrary rotations through Euler's formula (although one does not need Euler's formula to see this, of course). The other lesson to keep in mind here is that $e$ shows up whenever you solve linear ODEs. Abstractly this is because $e^{\lambda z}$ is an eigenvector for the derivative operator of eigenvalue $\lambda$.
I think these are important and fundamental questions, and it's a pity they aren't more clearly addressed anywhere in the undergraduate curriculum.
So, to summarize: $e^{it}$ is a complex number $\cos t + i \sin t$ which describes counterclockwise rotation by $t$ radians. It follows that if $z$ is a complex number of absolute value $1$, then the possible values of $\log z$ are the purely imaginary numbers $it$ such that $e^{it} = z$; in other words, they're the possible values of $t$ such that $z$ describes rotation by $t$ radians. So taking the logarithm of a rotation gives you an angle.
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