# 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?

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Your second paragraph is cool. Overall you really shouldn't think of the symbol $e$ in $e^{ix}$ as the number $e$. It will turn out to be it, but not from the get go. You should think of $e^{ix}$ merely as a concatenation of symbols which happens to include the symbol $e$ and which defined as $\color{grey}{e^{ix}:=}\cos (x)+i\sin (x)$. – Git Gud Jul 22 '13 at 19:28
A good way to think of Euler's gem is in terms of power series. $e$ comes naturally in the theory of infinite series, but so are $cos$ and $sin$. You can't think of replacing $e$ with something not related to series. – Torsten Hĕrculĕ Cärlemän Jul 22 '13 at 19:31
Excellent! Your second paragraph is just so Hermitian... And at the same time a partition function... and the clue to uncertainty principle... Absolutely beautiful! – al-Hwarizmi Jul 22 '13 at 20:09

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 $x=t,y=e^{-Mt}\cos (Rt),z=e^{-Mt}\sin (Rt)$.

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But that really leads to the question why do we define $a^x=e^{x\ln a}$... – Thomas Andrews Jul 22 '13 at 19:49
Is that a problem? I was just answering what the question of what happens when you put in other constants. I feel I did this. If curiosity leads on to a new question that's great! – Sharkos Jul 22 '13 at 19:52

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?

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By the way I tried this some weeks ago with the primes ($\log p$), you will get an amazing result... – al-Hwarizmi Jul 22 '13 at 20:12
@al-Hwarizmi This actually reminded me of another interesting problem showing density of a mapping of an irrational parameter onto the surface of a torus. I'm going to post it as a question. – Emily Jul 22 '13 at 20:15

Because $$e^{2\pi i \frac{k}{n}} \qquad 0 \le k < n$$ is the ($n$-th) root of unity.

One cannot neither make a perfect cyclotomy happen otherwise (see here>>>) nor what would require such a cyclotomy.

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Complex exponents can be think of the same way we do with real exponents. First we define that a^n = a*a*a... Then we ask what if n is any real number. Solution: limits. We can define exponential function

$$\ a^{x}=\lim_{n \to \infty}(1 + \frac{x}{n})^{n}$$ (visualizing this limiting process can helpful, but unfortunately I can't do it)

x can be time for example. We simply divide time smaller and smaller parts to calculate any real exponent we want. When x = 1, then

$$\ a = \lim_{n \to \infty}(1 + \frac{1}{n})^{n} = 2.71828... = e$$

Continuing this same analogy with complex numbers we define i^n = 90 degree rotation n times (-1 is 180 degree rotation and sqrt(-1) is half of that). We ask how to calculate with any real exponent n. Again, n (or x) can be time and the limiting process is defined similar:

$$\ a^{\imath x} = \lim_{n \to \infty}(1 + \frac{\imath x}{n})^{n}$$

When x = 1, then $$\ a^{\imath } = \lim_{n \to \infty}(1 + \frac{\imath x}{n})^{n} = {2.71828...}^{\imath} = e^{\imath}$$

It follows that x is in radians (in the unit circle arc lenght is simply x). In the above limiting process when we get closer enough then x is also the vertical cordinate of i and this small piece is also the piece of the unit cirle. So if we pick up x = $\pi$, then we divide this whole path infinitely many pieces and see that a piece is part of the unit circle. Complex exponentiation obey logarithmic behaviour (this can be seen many ways) so we can add all pieces together and conclude

$$\ e^{\imath \pi} = -1$$

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