# 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? – Emanuele Paolini Jan 31 '13 at 21:36
• – Rahul Jan 31 '13 at 21:38
• @manu-fatto The exponential function is simply e (2.718...) raised to a power. – caleb Jan 31 '13 at 21:38
• Oops. Yep. Fixed. – caleb 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. – caleb Feb 13 '17 at 16:36

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. – caleb 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 $e^{i\frac{3\pi}{2}}$ – Argon 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. – John Moeller 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. – Dider Sep 3 '15 at 20:45
• This is so great, simple and straightforward! :) I think that I finally get it intuitively, thank you! – Isti115 Mar 9 at 18:45

Converted from this article written for sci.math:

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 $(3)$.

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.

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 $2 \pi$.