Euler's identity: why is the $e$ in $e^{ix}$? What if it were some other constant like $2^{ix}$?

$e^{ix}$ 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.

Solution 1:

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?

Solution 2:

We in general define (ignoring $a=0$)

$$a^{ix}\equiv e^{i x\ln a}=e^{ixR -xM}$$

where we take the principal value of the logarithm and let $\ln a= R+iM$ be split into real and imaginary parts.

If your only aim is to have the locus being a circle then any $a$ such that $M=0$ will do - equivalently, you need $a>0$. The only difference from the $e$ case is that the speed at which you go round the circle is rescaled by $R$. (Your nice idea of differentiation and noticing orthogonality still works here.)

If you want the circle to be traversed at a speed such that $x$ is $2\pi$ periodic then you need $R=1\iff a=e$.

If you consider negative $a$ or general imaginary $a$ then you can see from the above formula that you get a circle multiplied by a new term $e^{-xM}$ which stretches it as it's drawn. This makes a (logarithmic) spiral.

In 3D space, the rescaled circles become helices which are more or less stretched out (like springs are compressed or elongated). The imaginary ones give various spirals stretched out across space. Graph them if you're interested by defining parametric equations $x=t,y=e^{-Mt}\cos (Rt),z=e^{-Mt}\sin (Rt)$.