# How does $e^{i x}$ produce rotation around the imaginary unit circle?

Euler’s formula states that $e^{i x} = \cos(x) + i \sin(x)$.

I can see from the MacLaurin Expansion that this is indeed true; however, I don’t intuitively understand how raising $e$ to the power of $ix$ produces rotation. Can anyone give me an intuitive understanding?

• What is your intuitive understaing of the exponential function? Jan 31 '13 at 21:36
• – user856
Jan 31 '13 at 21:38
• @manu-fatto The exponential function is simply e (2.718...) raised to a power. Jan 31 '13 at 21:38
• Oops. Yep. Fixed. Jan 31 '13 at 21:45
• I just happened across this video: youtube.com/watch?v=F_0yfvm0UoU, which helped give me a more intuitive understanding. Feb 13 '17 at 16:36

Starting with this formulation of $$e^x$$ $$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag{1}$$ and extending this definition to $$e^{ix}$$: $$e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n\tag{2}$$ For a complex number $$z$$, let $$|z|$$ be its magnitude and $$\arg(z)$$ be its angle. If it is not already known, only a small amount of algebra and trigonometry is needed to show that \begin{align} |wz|&=|w|\cdot|z|\tag{3a}\\ \arg(wz)&=\arg(w)+\arg(z)\tag{3b} \end{align} Induction then shows that \begin{align} |z^n|&=|z|^n\tag{4a}\\ \arg(z^n)&=n\arg(z)\tag{4b} \end{align} Let us take a closer look at $$1+\dfrac{ix}{n}$$. \begin{align} \left|\,1+\frac{ix}{n}\,\right|&=\sqrt{1+\frac{x^2}{n^2}}\tag{5a}\\ \tan\left(\arg\left(1+\frac{ix}{n}\right)\right)&=\frac xn\tag{5b} \end{align} Using $$(4a)$$, $$(5a)$$, and $$(2)$$, we get \begin{align} |e^{ix}| &=\lim_{n\to\infty}\left|\,1+\frac{ix}{n}\,\right|^n\\ &=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}\\ &=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n^2}{2n}}\\ &=\lim_{n\to\infty}e^{\frac{x^2}{2n}}\\[12pt] &=1\tag{6} \end{align} It can be shown that when $$x$$ is measured in radians $$\lim_{x\to0}\frac{\tan(x)}{x}=1\tag{7}$$ Using $$(4b)$$, $$(5b)$$, and $$(7)$$, we get \begin{align} \arg(e^{ix}) &=\lim_{n\to\infty}n\arg\left(1+\frac{ix}{n}\right)\\ &=\lim_{n\to\infty}n\arg\left(1+\frac{ix}{n}\right) \frac{\tan\left(\arg\left(1+\frac{ix}{n}\right)\right)}{\arg\left(1+\frac{ix} {n}\right)}\\ &=\lim_{n\to\infty}n\frac xn\\ &=x\tag{8} \end{align} Using $$(6)$$ and $$(8)$$, we get that $$e^{ix}$$ has magnitude $$1$$ and angle $$x$$. Thus, converting from polar coordinates: $$e^{ix} = \cos(x) + i\sin(x)\tag{9}$$ We get the rotational action from $$(9)$$ and $$(3)$$.

• Another way of showing $(6)$ is to note that for $n\ge\max\!\left(x^2,2\right)$, \begin{align} \left(1+\frac{x^2}{n^2}\right)^{n/2} &=\left(1-\frac{x^2}{n^2+x^2}\right)^{-n/2}\\ &\le\left(1-\frac{nx^2/2}{n^2+x^2}\right)^{-1}&&\text{(Bernoulli)}\\ &=1+\frac{nx^2/2}{n^2+x^2-nx^2/2}\\ &\le1+\frac{nx^2/2}{n^2/2}&&\left(n\ge x^2\right)\\ &=1+\frac{x^2}n \end{align}
– robjohn
Aug 21 '20 at 17:43

Consider a particle moving along the path $f(t)=e^{i t}$. It's instantaneous velocity is given by the derivative, and convince yourself that it is, treating $i$ as a constant, $ie^{it}$ Thus we see

$$\text{Velocity} = i\text{Position} = \text{Position (rotated by} \frac{\pi}{2} \text{radians)}$$

Because $f(0) = 1$, intitial velocity is $i$. Moving the position slightly and changing the velocity shows us that $|f(t)| = 1$ and thus $|\frac{d}{dt}f(t)|=1$. If $t =\theta$, the particle will have traveled $\theta$ radians around the unit circle.

• Ok, I think this helps me understand it. As you raise i to integer powers, it ends up rotating around the imaginary unit circle: $i^0=1$, $i^1=i$, $i^2=-1$, $i^3=-i$, and $i^4=1$. These positions (1, i, -1, -1) correspond the the following (x,y) positions: (1,0), (0,1), (-1,0), (0,-1). So it makes sense that multiplying the current position by i would result in a 90 degree ($\pi/2$) rotation. Jan 31 '13 at 22:35
• @caleb This is correct, yes. And the (principal) corresponding polar forms for these are, respectively, $e^{i\cdot 0}$, $e^{i\frac{\pi}{2}}$, $e^{i\pi}$ and $e^{i\frac{3\pi}{2}}$ Jan 31 '13 at 22:58
• I think this is the best, most succinct explanation of this phenomenon I've seen. Nice work. You could further expand this by pointing out that the acceleration is perpendicular to velocity, which will result in a circular trajectory. Feb 1 '13 at 0:28
• This is the exact explanation given in the beginning of the book "Visual Complex Analysis", there are deeper explanations later on which I recommend the original poster take a look at. Sep 3 '15 at 20:45
• This is so great, simple and straightforward! :) I think that I finally get it intuitively, thank you! Mar 9 '19 at 18:45

Let's look at the $|e^{ix}|$, this is always constant $(|\cos x + i \sin x| = \sqrt{ \cos^2x + \sin^2x} = 1)$. The only thing that is changed is $x$, now if we assign coordinates to real $(\cos x$) as $x$-coordinate and complex value $(\sin x)$ as $y$-coordinate (or imaginary axis), then this is same as parametric equation of unit circle with $x$ as parameter. As $x$ increases, the path traced by the point will be circular.

$\exp(z)$ is the function which is its own derivative. It's natural to introduce coordinates since we're thinking about the circle (a 2D figure).

We can consider it's real and imaginary parts: $\exp(iy) = c(y) + is(y)$, differentiating gives $\exp(iy) = - i c'(y) + s'(y)$ comparing with the previous gives $s'(y) = c(y)$ and $c'(y) = - s(y)$.

From the power series we find the real part is all the even powers so $c(-y) = c(y)$ and the imaginary part is all the odd powers so $s(-y) = -s(y)$, this lets us conclude Pythagoras' theorem $c(y)^2 + s(y)^2 = \exp(iy)\exp(-iy) = 1$.

From that we easily deduce that the path $(c(y),s(y))$ lies on the unit circle and is arc-length parametrized. Therefore it returns to its starting point when $y$ reaches $2 \pi$.

I don't know exactly what kind of intuition you're looking for. You're probably thinking about $e^{i\theta}$ the same way you would $2^2$; that is, $e^{i\theta}$ is found by multiplying $e$ by itself $i\theta$ times. While this is useful for introducing exponentiation of the form $n^m$ when $n,m$ are positive integers, it doesn't really make sense to try and apply this kind of reasoning to expressions of the form $a^x$ or $a^z$ for real or complex arguments.

The only sort of intuition I can suggest is the following: what is $e$? It's typically defined by the expression $\sum_{n=0}^{\infty} \frac{1}{n!}$. This isn't an alternate interesting fact but a definition for the number $e$. We also define $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, and we observe that this series converges for all real $x$. This is just what the expression $e^x$ means. Similarly, $e^{i \theta}$ is defined by $\sum_{n=0}^{\infty} \frac{(i \theta)^n}{n!}$, and it just so happens that this converges absolutely for all $z$, giving us Euler's formula.

This is why $e^{i\theta}$ is a rotation about the unit circle in $\mathbb{C}$. Because it's defined that way.

No rotation is involved. The equality is simply describing the location of a point on the complex circle. Motion implies time.There is no motion without time. Motion is change in the time dimension. Therefore, the motion of a point in the complex plane is, in itself, a 3 dimensional graph. We can involve time by changing the equation to $$e^{i\Theta} = \cos(t)+i\sin(t)$$. This gives 3 dimensions, time and the complex circle. A perfect example is a point moving on a circle on an oscilloscope when the horizontal and vertical deflection voltages are $$V_h=V\cos(t), V_v=V\sin(t)$$. Deflection of the electron beam is proportional to voltage. $$e^{i\theta}=e^{it}$$. This easy substitution just begs to be motion before changing $$\theta$$ to $$t$$.

We could "freeze" time and change the graph to another form by converting time to another dimension. Call it the $$z'$$ dimension, where $$z'=t$$. The graph becomes the continuous function $$z'=f(x)=e^{ix}+x$$, where $$x$$ (or substitute $$t$$) is a real number. It is a helix in the perpendicular $$z'$$ direction.

"Freezing" time and changing dimensions is commonly done. Throw a ball straight up in the air (the $$y'$$ direction). Convert the time dimension to the orthogonal $$x'$$ dimension. Graphing results in a parabola in the $$(x',y')$$ plane, where $$x'=t'$$.

It is worth repeating that spinning the point or throwing the ball sideways (the trajectory of a ball thrown laterally is a parabola) is so compelling due to the easy substitutions. Remember, $$t$$ is frozen. Any additional motion would be yet another dimension from the converted graphs, which are geometric entities in their own right. Consider $$f(x)=x$$. Most people would consider this a static function. Few people would conceive of a point moving along the line. A point used as a cursor would be just fine. But formalizing this motion would result in going into another dimension. Resist temptation. But if you have a quarter and must...

Treat yourself and go to a spiral gumball machine to watch a 4 dimensional gumball. It is also the graph of $$f(z)=re^\frac{ia_rt^2}{2}+\frac{ka_rt^2}{2}$$, where z is a complex number. The solution is wherever the ball is at a given time.

$$z'=\frac{ka_rt^2}{2}$$$$t'=re^{\frac{ia_rt^2}{2}}$$$$f(z)=z'+t'$$
The primed numbers are used to differentiate coordinates from other numbers
r is the radius of the gumball drop
$$a_r$$ is the acceleration of the ball down the slope, different for different helices
k is a proportionality constant related to the pitch
For constant a, $$d=\frac{at^2}{2}$$ or $$s=\frac{at^2}{2}$$ for circular units

In general, consider the complex plane to be the initial location of the gumball for any size gumball machine with helices on the same helicoid ("filled in" helix or a screw).

The big picture is the important thing. Every point on the complex plane $$(r,\theta)$$ is mapped by a function $$f(z)$$ to 2 new variables $$(z',t')$$.

I visualize this as a rotating plane accelerating on the $$z'$$-axis changing shape as the inner radii accelerate faster.

Going back to the beginning, modify Euler's formula to: $$e^{if(\theta)}=cosf(\theta)+ isinf(\theta)$$
The questioner's implicit choice is $$t=f(\theta)$$. Euler's formula is $$\theta=f(\theta)$$. Euler's case is static. But, time can exist w/o motion. (A circle just sitting there is actually, in space-time, a cylinder). An analogy would be $$f(x)=x$$. Adding $$z'$$ changes the line to a plane. $$f(x)=x$$ is so fundamental it doesn't need 2 dimensions. It can be described by the real number line as $$x=x$$. Replacing $$f(\theta)$$ with $$t$$ or any function of $$t$$, $$f(t)=f(\theta)$$. Any non-trivial $$f(t)$$ implies another dimension $$t'$$, (which means it can be graphed uniquely). Motion is described by any function $$f(t)=f(\theta)$$, where $$t=\theta$$ is a real number. This non-trivial $$f(t)=f(\theta)$$ is what sends the gumball into the 4th dimension as a complex function. The graph of $$e^{it}$$ is "smeared" by multiple values of the phasor representing it's location. In most cases, this is resolved by creating a $$z'=t'$$ dimension, as noted earlier, to graph the function. In the gumball's case, the $$z'$$ dimension is already taken up by the necessity of the gumball to stay in the fixed loci of the helix. The only choice is to go into the time domain to graph the function uniquely. Velocity, in a circle, IS ORTHOGANAL to that circle, confirming motion is in a new dimension.

$$e^{i\theta}$$is more than position.

It is the time dimension.

Suppose we use the following definition of the exponential function: $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n \, .$$ This definition makes sense even when $$x$$ is a complex number. If $$n$$ is very large, then $$1+\frac{i\theta}{n}\approx\cos\left(\frac{\theta}{n}\right)+i\sin\left(\frac{\theta}{n}\right)$$ using linear approximation. Hence, $$\left(1+\frac{i\theta}{n}\right)^n\approx\cos\theta+i\sin\theta \, .$$ by de Moivre's theorem. It seems reasonable that in the limit as $$n\to\infty$$, this approximation turns into exact equality.

• Could the person who downvoted please explain why they have downvoted?
– Joe
Aug 22 '21 at 20:48