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BloomLocke367
 one year ago
Tutorial: Understanding imaginary and complex numbers.
BloomLocke367
 one year ago
Tutorial: Understanding imaginary and complex numbers.

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BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.2Imaginary 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.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.2An 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 \(\large i^2=(\sqrt{1})^2=1\).

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.0this post is 13 days old...

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.2A 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here's some stuff you missed out: if we take the ring of real polynomials in the indeterminate \(X\), and we quotient out the twosided ideal generated by \(X^2+1\) so that, essentially, \(X^2+1\equiv 0\), we get something equivalent to the complex numbers: $$\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}$$

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.2Umm I'm not done yet

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.2I just haven't been able to continue.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0additionally, "the" complex numbers are the only twodimensional real algebra that make a field; the splitcomplex 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 \((a,b)\cdot(c,d)=(acbd,ad+bc)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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 \((a,b)\mapsto b\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because 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 \(i^2=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in general, all such twodimensional 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 splitcomplex numbers
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