expansion of complex numbers.

[perusal]

a,b ∈ [MATH] \mathbb C[/MATH]
|a + b|[MATH]^2[/MATH] = |a|[MATH]^2 + |b|^2 +\bar ab\ + a\bar b [/MATH]
Having trouble with this one.

Anyone care to walk me through it?

Also on a side note, is the square of a complex number necessarily positive. I'm thinking it isn't but I don't have a proof or anything. Is there any way to order imaginary numbers? is i>2i ?

 

[Dr.Peterson]

Are you asking how to prove this is true for all a and b? The words in a problem are as important as the symbols!

Think about the fact that

[MATH]|a + b|^2 = (a+b)\overline{(a+b)}[/MATH]​

As for being positive, that concept only applies to real numbers, and the square of a complex number is not always real. So the question doesn't even have any meaning.

Similarly, there is no way to say that one complex number in general is greater than another, so ">" does not apply. You could, I suppose, define it for pure imaginary numbers, but that won't interact in a useful way with any other concepts. Is there some situation in which you think it would be meaningful?

 

[perusal]

Thank you

 

[HallsofIvy]

A "field" (an algebraic construct in which you can both add and multiply the "numbers" and every number except 0 has a multiplicative inverse) is an "ordered field" if there is a an ordering (a< b) such that a
1) For any a and b, one and only one is true: a< b or b< a or a= b.
2) if a< b then a+ c< b+ c.
3) if a< b and 0< c then ac< bc.

If there were such an order on the set of complex numbers, then, since i is clearly not 0, either i< 0 or i> 0. If i> 0 then i(i)= i^2= -1> 0. This doesn't have to be an extension of the usual order on the real number so "-1> 0" is not immediately a contradiction. However, if -1> 0 then -i>0 so that 0> i, contradicting our assumption that i> 0. If i< 0 then i(i)= i^2= -1< 0 but then -i< 0 so that i> 0, again a contradiction. There is no way of defining an order on the complex numbers consistent with its field properties.