$e^{ix}$ describes a unit circle in polar coordinates on the complex plane, where x is the angle (in radians) counterclockwise of the positive real axis.
My intuition behind this is that $\frac{d}{dx}e^{ix}=i\cdot e^{ix}$. Since multiplication by i is a 90-degree rotation, we could think of $e^{ix}$ as the position vector of a particle and $\frac{d}{dx}e^{ix} = i\cdot e^{ix} $ as its velocity (x could be time). The velocity is always perpendicular to the position vector, so we have circular motion.
Hopefully I've described this well, see also
http://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/ if you don't understand where I'm coming from.
What I don't understand is why you need the "e" in Euler's identity. What if it were some other constant: for example, 2? You wouldn't get a circle, but how can I visualize what it is that you would get?
For example, what would $2^{ix}$ look like on the complex plane? I note that $2^{ix} = e^{ix\cdot ln(2)}$, and we could substitute that into Euler's identity and get $e^{ix\cdot ln(2)}=cos(x\cdot ln2) + i\cdot sin(x\cdot ln2)$.
So my question really has two related parts:
1) Why do we take e (and not some other number) to the power of ix to get a circle?
2) What would it look like if we took some other number to the power of ix? $e^{ix}$ really gives us a constant-radius spiral in three dimensions (e.g. http://www.songho.ca/math/euler/euler.html), what would $2^{ix}$ look like in complex 3d space? How could I have figured that out?
Thank you for your help.
Best Answer
Great question! Let's plot it for $2^{ix}$, with $0 \le x \le 2\pi$ (as we would with $e^{ix}$), and see what happens:
Woah, not quite a circle.
Why is this? Well, because of course our frequency is $\log 2$! For $0 \le x < 2\pi$, we don't quite make it all the way around. How far would we need to go? (It's not that hard to figure out).
To answer your first question, the reason we take $e$ is because it's the only number that jives with our notion of $2\pi$ radians in a circle.
For your second question, it means you'd get the same thing. But with a tighter/looser spiral!
Edit: Since $2\pi$ radians isn't enough to get us a full circle with $2^{ix}$, maybe we could figure out how many "log-two-dians" are necessary.
Alternatively, suppose we want to do things in degrees without an explicit conversion to radians. What would our base need to be?