Question

Understanding Imaginary and Complex Numbers Tutorial
Creator of Tutorial: Bloomlocke367 and Unknown

Answer 1

Imaginary and Complex numbers may be confusing when you first see them, but they're actually not that bad. You just have to understand what they mean.

Answer 2

An imaginary number, \(i\), simply represents \(\sqrt{-1}\).

In the past, you may have been told you can't take the square root of a negative number, but imaginary numbers allow you to do so.

You should also understand that \(\large i^2=(\sqrt{-1})^2=-1\).

Answer 3

A complex number is made up of two components. The real component, and the nonreal component. It is expressed in the form of \(a+bi\), where a is the real number, and bi is nonreal.

Answer 4

UNKNOWN: Here's some stuff that was left out: if we take the ring of real polynomials in the indeterminate \(X\), and we quotient out the two-sided ideal generated by \(X^2+1\) so that, essentially, \(X^2+1\equiv 0\), we get something equivalent to the complex numbers: \[\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}\]

Answer 5

UNKNOWN: additionally, "the" complex numbers are the only two-dimensional real algebra that make a field; the split-complex and dual numbers both have zero divisors. it is also equivalent to taking \(\mathbb{R}^2=\{(a,b):a,b\in\mathbb{R}\}\) and equipping it with a natural commutative, bilinear product such that: \[(1,0)\cdot(1,0)=(1,0)\\(1,0)\cdot(0,1)=(0,1)\\(0,1)\cdot(0,1)=(-1,0)\]this in turn generates the general form \((a,b)\cdot(c,d)=(ac-bd,ad+bc)\)

Answer 6

UNKNOWN: the projection onto the first element is denoted \(\Re :\mathbb{C}\to\mathbb{R}\) such that \((a,b)\mapsto a\), so it follows that \(\{(a,0):a\in\mathbb{R}\}\) is just a copy of the real line. the second projection is \(\Im:\mathbb{C}\to\mathbb{R}\) so \((a,b)\mapsto b\)

Answer 7

UNKNOWN: because of the natural identification of \(\{(a,0)\}\) with the real line, and the fact that multiplication restricts nicely to this line to match the structure of \(\mathbb R\),we can identify \((1,0)\) with \(1\) and we can identify the second basis element \((0,1)\) with \(i\), so our elements look like \((a,b)=a(1,0)+b(0,1)=a+bi\). now the laws above look much more simple: \[1\cdot 1=1\\1\cdot i=i\\i\cdot i=-1\]so it follows \(i^2=-1\), and this multiplication amounts to taking a field extension of \(\mathbb{R}(i)\) with an element \(i\) defined so that \(i^2=-1\)

Answer 8

UNKNOWN: in general, all such two-dimensional real algebras come from \(\mathbb{R}[x]/(X^2+k)\) -- in the case \(k>0\) we get something isomorphic to the complex numbers, i the case \(k=0\) we get the dual numbers, and in the case \(k<0\) we have the split-complex numbers

Answer 9

thanks