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

  • $\begingroup$ 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. $\endgroup$ – Please Delete Account Mar 14 '11 at 23:54

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.

  • $\begingroup$ This explains something about $e$ and exponeniation. What about the inverse, the natural log? $\endgroup$ – Mitch Mar 15 '11 at 0:32
  • 1
    $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Mar 15 '11 at 1:02
  • $\begingroup$ 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. $\endgroup$ – Mitch Mar 15 '11 at 1:57
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    $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Mar 15 '11 at 2:01

When I personally went on the quest for an intuitive grasp of the relationship between Euler's Number and Rotations. I would almost invariably find explanations that missed the point entirely and instead simply explained the motivation for imaginary exponents in general. Yes this is required knowledge but it's not the end of the story. As the picture below illustrates any Real Number to an Imaginary Exponent will create a Parametrization of a Unit Circle in the Complex Plane.

The real reason we choose to use Euler's Number over any other is because it is the unique "Growth Factor" that correlates a Unit Circle Parametrization with Arc Length.

That is to say, for $B^{i t}$, the Arc Length of a Unit Circle = 1 when t = 1, if and only if $B = e$.

So $e^{it}$ is a pretty nice and convenient little formula with the Parameter $t$ corresponding exactly to our measure of Arc Length. From which we also indicate Angle in general through Radians.

However we are again left with another question now. Why exactly is it that Euler's Number is the "Growth Factor" that correlates a Unit Circle Parametrization with Arc Length?

After all $e$ usually presented in a purely analytic fashion, related to function growth and certainly doesn't seem to have such direct connection to the Arc Length of a Circle.

This is really the crux of the matter and is something that I think can only really be seen by examining the "Moving Particle" analogy in detail.

Warning: Wordy explanation ahead!

Part 1:

I find that thinking about this in terms of the Physics is the clearest and most complete explanation.

Interestingly enough it shows how the so-called "deep connection" between Euler's Number and Pi turns out to be nothing more then the familiar property of the Natural Exponential Function being equal to all of its Derivatives.

enter image description here

There's a lot going on in the above three pictures so let me explain them point by point.

Each graph represents a Unit Circle Parametrization in the Complex Plane with Real Numbers going horizontally and Imaginary Numbers vertically.

The first graph uses the equation $f(z) = 2^{i t}$. The second $f(z) = e^{i t}$ and the third $f(z) = 3^{i t}$.

Where $t$ runs from $[0,1]$.

For each graph the Light Blue Curve corresponds to function itself while the Yellow, Green and Orange Curves are the 1st, 2nd and 3rd Derivatives.

The correspondingly colored Vectors stemming from the Origin are the Position Vectors that "sweep out" their respective Curve.

The Light Blue Position Vector (which is the function itself, not one of its Derivatives) also has additional Vectors stemming from it which indicate its Velocity (Dark Blue), Acceleration (Red) and Jolt (Purple). Where Jolt is simply the Derivative of the Acceleration.

Notice how in all three different bases the Velocity, Acceleration and Jolt Vectors are always Perpendicular to each other.

However only the formula using Euler's Number as a Base has Derivatives that seem to "line-up" with each other.

The understand why this is and why it's important we'll have to go full on Physics mode.

Part 2

First, assume the Unit Circle Parameter is Time in Seconds.

The essential idea is that in order for a Radius of Length 1 to move 1 Arc Length in 1 Second it is required to have a Velocity of 1, Acceleration of 1, Jolt of 1, etc.

(As well as all being Perpendicular to each other, but any Growth Factor of a General Exponential Function gives the Perpendicular Derivatives property.)

If we interpret Arc Length as a general measure of Distance.

In order for a Particle to move 1 Meter in 1 Second it is required to have a Velocity of 1 Meter Per Second.

For this to hold true indefinitely, it is required to have a Parallel Acceleration of zero. That is to say, it can neither slow down nor speed up.

However it may have Perpendicular Acceleration. This Perpendicular Acceleration should preserve the above properties no matter the magnitude as long as it is only ever strictly perpendicular to the Velocity.

If we now add an additional constraint, that of a "Stable Orbit" of the form of perfect eternal circular motion.

Then we find that the Perpendicular Acceleration should change the Velocity by 1 Meter Per Second every Second (given an Orbit with a Radius of 1 Meter).

This is because it must always be pointing "inwards" but which direction "inwards" corresponds to constantly changes as the particle orbits.

Consequently we find that none of the Particle's Derivatives (up to the Infinitith Order Derivative!) can be Zero and in fact must all be Constant. In this situation in fact they must all be 1. (That is to say, of course, that all their Magnitudes must be Constant and in this case are equal to 1. Their Directions are constantly spinning around to maintain their positions perpendicular to each other.)

This is because in eternal circular motion all of a Particle's Derivatives must be constantly "correcting" the Derivatives lower than it in order to maintain the constant instantaneous changes in motion forever.

I hope this rambling explanation was helpful. I only post it because after spending weeks reading about Euler's Number, Trigonometry and Complex Numbers this was the only explanation that finally made all the details "click" for me and I couldn't find such an explanation on-line anywhere after much searching.


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.


$ \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.


How is the natural logarithm related so closely to circles ?

The natural logarithm is $\ln x=\displaystyle\int_1^x\frac1t~dt.~$ Now, $f(t)=\dfrac1t~$ is a hyperbola, which, when

rotated by an angle of $45^\circ,$ and scaled by a factor of $\sqrt2,~$ yields $x^2-y^2=1,$ which looks

remarkably similar to the equation of the unit circle, $x^2+y^2=1,$ whose constant is $\pi.$

Substituting y by iy in the former equation will yield the latter expression, and vice-versa.

Also, circles and hyperbolas are both conic sections. So it should come as no surprise if one

were to find a formula involving Euler's constant e and imaginary numbers, ultimately

linking the two famous types of geometric shapes, known and studied since ancient times.

I also recommend the following related questions:


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