Are there any numbers other than Complex Numbers which contain something more that Real and Imaginary numbers?
[Math] Are there any numbers more fundamental than Complex numbers
complex numberssoft-question
Related Solutions
Actually a plethora of engineering domains use complex numbers (in circuits, mechanics, oscillations etc..) for example phasors
The reason this is done is because complex numbers by De Moivre's theorem relate nicely to periodic signals and systems, and treated using operations of multiplication/addition etc..
Also many trigonometric formulas are simplified if expressed in complex form (related to what is stated before).
These are indeed examples of real-world applications of complex numbers.
UPDATE: Another example of real-life complex arithmetic is clock-like arithmetic (sth similar to modulo arithmetic for integers). Meaning numbers of the form $e^{ia}$ where a is a real number. These numbers consitute a group ($U(1)$) the unitary group of 1 parameter and is also a Lie group. One can think of it as a counterpart of $\mathbb{Z_p}$ modular group but with real numbers.
NOTE A realist/constructivist approach to mathematcal entities need not be constrained in natural numbers only. Given interpretations (e.g as a process) indeed realist representations of these entities are not only possible but realizable as well.
One way to think of the complex plane $\mathbb{C}$ is to think of it as the real coordinate (or $xy$-) plane $\mathbb{R}^2$. In $\mathbb{R}^2$, we have two real-numbered axes (in particular, the $x$ and $y$ axes). An element of $\mathbb{R}^2$ is an ordered pair $(x, y)$, where $x$ and $y$ range over all the real numbers $\mathbb{R}$.
What I am about to say is not exactly rigorous, but I purposefully omit rigor in favor of your understanding.
Now, what happens if we decide to multiply $y$ by $i$, for all $y$ in $\mathbb{R}$? We obtain $i \mathbb{R}$, where each element is of the form of $i y$, where $i$ is of course defined as usual and $y$ is a real number. Consequently our vertical $y$-axis now becomes imaginary, and we obtain the complex plane $\mathbb{C}$. Note that just like the real plane $\mathbb{R}^2$, $\mathbb{C}$ is 2-dimensional. To recap, $\mathbb{C}$ has two axes, one real axis and one imaginary axis.
What does an element of $\mathbb{C}$ look like? An element $z$ of $\mathbb{C}$ has the form $a + bi$, where $a$ and $b$ are real numbers; we call $z$ a complex number. If we were to graph a complex number $z = a + bi$, we would graph it just like we would an ordered pair $(a, b)$ in $\mathbb{R}^2$. E.g., the number $z = 1 + i$ would correspond to the point $(1, 1)$; the number $z = 1$ would correspond to the point $(1, 0)$. This second example demonstrates that a real number is in fact a complex number as well.
I am not sure if you have seen linear algebra or not, but your statement in the second sentence of your first paragraph can be explained with such ideas. In the real plane $\mathbb{R}^2$, we have two directions; namely, the $x$ and $y$ directions. It turns out that we only need the number $x = 1$ to generate the entire $x$-axis. Similarly we may do the same with the $y$-axis. Together, we obtain the whole plane $\mathbb{R}^2$. $\mathbb{C}$ carries the same idea, except your vertical axis is generated by $i = 1\cdot i$ instead of just $1$.
To further address your question, recall $i$ is defined as $i = \sqrt{-1}$. Thus, $i^2 = -1$, $i^3 = -i$ and $i^4 = 1$. What have we accomplished here? We have obtained directions for the vertical and horizontal axes. From here we may generate all of $\mathbb{C}$.
I am not entirely sure how to answer your next few questions, but recall from above that $\mathbb{C}$ is 2-dimensional. Thus, $\mathbb{C}^2 = \mathbb{C} \times \mathbb{C}$ would be 4-dimensional. Generalizing this notion, $\mathbb{C}^n$ is $2n$-dimensional. Hence I do not think there is really an analogy to the $z$ axis if we are referring to the third axis in $\mathbb{R}^3$. The best analogy I can come up with are several-dimensional complex number systems such as $\mathbb{C}^2$.
Best Answer
Yes. There is in fact a well-structured set of numbers of dimension $2^n$ for all $n$, but after a certain point it simply becomes pointless and unwieldy. The next set of numbers is called the quaternions of the form $a+bi+cj+dk$, where $a,b,c,d\in\mathbb{R}$ and $i,j,k$ are all square roots of $-1$. In the quaternions, $ij=k, jk=i, ki=j$ but $ji=-k, kj=-i, ik=-j$ (that's right, they're non-commutative). The complex numbers $\mathbb{C}$, though, are considered more fundamental than the quaternions $\mathbb{H}$ or the sets that follow (the octonions, sedenions, etc.), since the complex numbers are an abelian field with algebraic closure (algebraic closure of a field $F$ means that all polynomials $\sum a_ix^i$ with $a_i\in F$ have all of their roots in $F$. This cannot be said of $\mathbb{R}$, since $x^2+1$ has real coefficients but imaginary roots.), giving it the greatest structure of all of these number sets. Funny things start to happen after the quaternions, as the octonions are no longer associative and the sedenions can have zero divisors (which means two non-zero elements $a$ and $b$ can have the property $ab=0$).