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As an engineer, I learned a lot about how to use complex numbers. One way I have heard $i$, the unit complex number, defined is:

It is orthogonal to the real number line. Because $\frac{\mathrm{d}}{\mathrm{d}x} e^{x} = e^{x}$, we can clearly see $e^{\pi i} = -1.$

and the professor draws the unit circle.

However, this is not a unique definition of $i$. Whereas real 1 is $\hat{x}$ and $i$ represents $\hat{y}$ on the whiteboard, there are $\hat{z}$, $\hat{w}$, etc. which also satisfy the equations above.

We can use rotation to translate $\hat{z}$ to $\hat{y}$. Therefore it is still possible to use the "two-dimensional" definition $\mathbb{C}(a) <=> \hat{x} \Re(a) + \hat{y}(-\Im^2(a))$. In other words, if only care that $i^2 = -1$, then it does not matter which from the infinite set of $i$s you choose.

However we could consider $?(a) <=> \hat{x} \bullet a + \hat{y} \bullet a + \hat{z} \bullet a$ which satisfies $e^{\pi \hat{y}} = -\hat{x}$, $e^{\pi \hat{z}} = -\hat{x}$, and $e^{\pi \hat{y}\hat{z}} = -\hat{x}$.

Are these considered numbers in any way more useful than the normal complex numbers?

Note: I am not asking about $\mathbb{C}^3$, $\mathbb{C}^4$, ... as discussed at 3 Dimensional Complex Plane?

William Entriken
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    Have you read about quaternions or octonions? – costrom Jan 22 '16 at 17:32
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    It really depends on what you mean, but the natural place to start is the quaternions that costrom mentions. These naturally form a 4-dimensional space, and in short there is no way to build a 3-dimensional space that satisfies certain axioms that are natural to ask for (see, e.g., http://math.stackexchange.com/questions/911720/why-should-i-consider-the-components-j2-and-k2-to-be-1-in-the-search-f/911809#911809 ). – Travis Willse Jan 22 '16 at 17:34
  • It turns out there isn't much you can do in three dimension. In four dimension (quaternions,) you lose commutativity of multiplication. In eight dimensions (octonions,) you lose associativity of multiplication. (It, of course, depends on what you want from such an extension - you can define lots of rings that have zero divisors in three dimensions, for example.) – Thomas Andrews Jan 22 '16 at 17:44
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    What does $\frac{d}{dx}e^{\pi i}=e^{\pi i}$ mean? That's terrible notation. If $f(x)=e^{x}$ then $f'(i\pi)=f(i\pi)$. But what if $f(x)=e^{ix}$? Then $f'(\pi)=if(\pi)$. – Thomas Andrews Jan 22 '16 at 17:46
  • @ThomasAndrews thank you, I have corrected this notation – William Entriken Jan 22 '16 at 19:59

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William Rowan Hamilton introduced quaternions in the 19th century. These generalize complex numbers and also generalize cross-products of vectors. A quaternion is an object of the form $$ a + bi + cj + dk \tag A $$ where $a,b,c,d$ are real numbers and $i,j,k$ are objects that can be multiplied as follows: \begin{align} i^2 = j^2 = k^2 & = -1 \\ ij = k & \qquad ji = -k \\ jk = i & \qquad kj = -i \\ ki = j & \qquad ik = -j \end{align} Addition and subtraction of quaternions is term by term.

Notice that $\pm i, \pm j, \pm k$ are not the only square roots of $-1$: if $a=0$ and $b^2+c^2+d^2=1$, then the square of the expression $(\mathbf A)$ above is $-1$.

The use of quaternions in physics was superseded by the use of vectors in $\mathbb R^3$ with the usual dot- and cross-products, but quaternions are used today in computer graphics.

Quaternions afford an easy way to see that the space of rotations of $\mathbb R^3$ that leave the origin fixed is not simply connected, as follows. First show that the map $$ bi+cj+dk \mapsto (Pi + Qj +Rk)\Big(bi+cj+dk\Big)(Pi + Qj +Rk)^{-1} $$ is a rotation of the $3$-dimensional space of "pure" quaternions ("pure" means the real part $a$ is $0$). Then observe that the two quaternions $\pm(Pi + Qj +Rk)$ both represent the same rotation. That means a path on the sphere from $Pi + Qj +Rk$ to $-(Pi + Qj +Rk)$ corresponds to a path in the space of rotations, from one particular rotation to itself, that cannot be contracted to a point.

Michael Hardy
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    I read this part a few times "Quaternions afford an easy way to see that the space of rotations of $\mathbb R^3$ that leave the origin fixed is not simply connected". This part is very interesting and is very interesting to see why you would use them. Thank you. – William Entriken Apr 15 '16 at 22:02