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$e^{i \pi} = -1$. 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 $e^{ix} = \cos(x) + i\sin(x)$?

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FWIW the moving particle argument is presented in the book Visual Complex Analysis by Tristan Needham. I would recommend a reading to anyone interested. It is an excellent book, but without any disrespect to the author, I find the answer below to be of higher quality than the book's. –  Please Delete Account Mar 14 '11 at 23:54

3 Answers 3

up vote 14 down vote accepted

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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This explains something about $e$ and exponeniation. What about the inverse, the natural log? –  Mitch Mar 15 '11 at 0:32
@Mitch: I'm not sure I understand the question. Take everything I said about $e$ and take the natural log of both sides. If you accept that $e^z$ is related to rotation, then its inverse is automatically also related to rotation. –  Qiaochu Yuan Mar 15 '11 at 1:02
I hope I'm not being obtuse, but I just don't see immediately what the composition inverse of exponentiation (the natural log) corresponds to here. Is it rotation in the other direction? That seems to be the multiplicative inverse (the negative angle). What corresponds to taking the log of ...well of what? The log of a complex number is an angle? That's what isn't obvious. –  Mitch Mar 15 '11 at 1:57
@Mitch: I have added a remark that I hope clarifies this. You are taking the logarithm of a complex number that represents a rotation and getting the angle of the rotation. –  Qiaochu Yuan Mar 15 '11 at 2:01

The complex logarithm is quite different from the real-valued logarithm. In fact, it is multi-valued.


While it agrees with the real-valued logarithm on the real axis, if you take $R>0$, then as you traverse the circle $|z| = R$, the value of the complex logarithm increases linearly with the $\arg z$ until you return to the same place. For example, if you take the unit circle, start at $1$, pass through $i$, $-1$, and $-i$, then when you return to $1$ the logarithm will have increased by $2\pi i$. This comes from the fact that $\exp$ is periodic, as you have noted by realizing its relationship to Euler's identity.

Typically, we take one branch of the complex logarithm (which has discontinuities along a ray from the origin, whose location we can choose, and we typically choose either the positive or negative real axis), but the actual function looks a bit like a helix or a spiral.

Therefore, the relationship between $\log$ and the circle is that, each time you go once around the circle, you gain $2\pi i$.

If you are interested in more of this, you should study holomorphic functions and residues. In a region where a holomorphic function is holomorphic (with certain nice properties like being simply connected), the integral around any closed curve is zero. Simple poles of a holomorphic function $f$ are points where, locally, $f$ looks like $1/z$. You will recall that, in real analysis, the anti-derivative of $1/z$ is $\log(z)$. Since $\log$ has a branch cut, this fails to be a holomorphic anti-derivative of $1/z$, so you need some more power to integrate $1/z$, but it turns out that the integral along a circle around a pole like that will be exactly $2\pi i$, and this is precisely because, by taking one branch of $\log$, you get a jump of $2\pi i$ when you return to your starting point.

More simply, recall that multiplication is the same as addition of logarithms. In the complex numbers, multiplication corresponds to addition of arguments (angles). Therefore, you can add something ($2\pi i$ it turns out) to your logarithm, which corresponds to multiplying by something that takes you once around the circle. This is maybe an easier to see, more intuitive relationship.

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$ \ln{\left( x+ \hat{i}\,y \right)} = \ln{\sqrt{\left( x^2+y^2 \right)}+ \hat{i}\,\tan^{-1}\left(\dfrac{y}{x}\right)} $

when polar coordinates are used $x=r\,\cos\theta$ , $y=r\,\sin\theta$

$ \ln{\left( r\,\cos\theta+ \hat{i}\, r\,\sin\theta \right)} = \ln{r} + \hat{i}\,\theta $

So the logarithm of an complex number represents the rotation angle in the imaginary units.

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