Is there such thing as an imaginary (imaginary number)?

Question

In other words... is there such a thing that is to imaginary numbers what imaginary numbers are to real numbers? And could this be expressed as a "complex" type number? If a complex number is in the form x + yi, I guess this would be in the form of x + yi + zj? Does that exist as a concept?

Answer 1

Yes. Numbers akin to perhaps what you have suggested do exist. The Quaternions are a canonical example. Your basic old complex number can be written $a + ib$. Here we have numbers of the form $a + ib + jc + kd$. The rules for multiplying such numbers are discussed on the Wikipedia link above.

These have applications in applied math, especially things requiring three dimensions. Pure maths also uses them sometimes as they form a skew field.

Answer 2

In a sense, yes. There is a number system called the Quaternions, which is an extension of the complex numbers. You can find the Wikipedia page here: http://en.wikipedia.org/wiki/Quaternion

In the complex numbers, we have $i^2 = -1$, but in the quaternions, which extend to $i, j, k$, we still have $i^2 = -1$, but we also have different rules such as $ij = k$ and $jk = i$. This number system plays a large role in simplifying engineering equations, just as complex numbers simplify equations describing electromagnetism and fluid dynamics.

Certainly, you can extend this system to arbitrarily many variables, so long as the basic rules (axioms) are consistent. And really, that's something mathematicians do: Create structures, give it basic rules, and see what comes out of that.

Answer 3

No there is no such thing. To understand this you have to understand the imaginary unit: $i$ solves the equation $x^2 = -1$--that is $i^2 = -1$. Right away there are is a "problem" because $-i$ solves this equation just as well as $+i$: $\left(-i\right)^2 = (-1)^2i^2 = +1i^2 = -1$ and $\left(+i\right)^2 = (+1)^2i^2 = i^2 = -1$ so which do we choose? Do we choose $-i$ or do we choose $+i$. The answer is we choose neither and instead define $i$--we define $i$ such that $i^2 = -1$--it doesn't matter if we really mean $-i$ or $+i$--we just choose one of those two (we "choose" $+i$ because why carry around the burden of the negative when it's equivalent).

So to say there is an analogue to the complex unit $i$ would mean that there is some value of $x, y \in \mathbb{C}$ such that the following equation cannot be solved with complex numbers such that $x^2 = y$ where $x,y \in \mathbb{C}$. This is not possible, there is a value $x \in \mathbb{C}$ such that $x^2 = y$ for all values in $y\in\mathbb{C}$. We can go further and say that there exists at least one value $x \in \mathbb{C}$ such that for all values of $y \in \mathbb{C}$, $x^n = y$ where $n$ is a natural number (and you could expand to $n$ being a positive real value or a non-zero complex value).

I'm not familiar with these Qauternion values but these sound like higher dimension analogues. I am strictly talking about $1$D values. The imaginary number solves an insolvable problem from $\mathbb{R}^1$. Once we extend $\mathbb{R}^1$ to include $\mathbb{C}^1$ all such problems become solvable.

Answer 4

Right. For some flavor, the complex numbers are perfectly represented by matrices of this pattern: $$ \left( \begin{array}{rr} a & b \\ -b & a \end{array} \right) , $$ with $a,b \in \mathbb R.$ In particular, $$ 1 \rightarrow \left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right) , \; \; \; \; i \rightarrow \left( \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right) . $$

Once you accept the complex numbers, the quaternions are perfectly represented by matrices of this related pattern: $$ \left( \begin{array}{rr} \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{array} \right) , $$ with $\alpha, \beta \in \mathbb C.$ In particular, $$ 1 \rightarrow \left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right) , \; \; \; \; i \rightarrow \left( \begin{array}{rr} i & 0 \\ 0 & -i \end{array} \right) , \; \; \; \; j \rightarrow \left( \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right) , \; \; \; \; k \rightarrow \left( \begin{array}{rr} 0 & i \\ i & 0 \end{array} \right) . $$ If you prefer, you can replace the 2 by 2 complex matrices with 4 by 4 real matrices, whereupon all the entries in the matrices for $1,i,j,k$ are $0,\pm 1.$ Just a bit harder to remember.