Re: A89: Re: Re: Re: Re: Re: C question... actually two C questions...


[Prev][Next][Index][Thread]

Re: A89: Re: Re: Re: Re: Re: C question... actually two C questions...




Mmmmmm, more math...DeMoirvre's theorem (to find all complex roots of a #). 
Can't  remember it, though (been WAAAAAAY to long since Pre-Cal). I remember 
it had to do with trigonometric representations of complex #'s, and you 
repeated it n-1 times (n being the root taken). Now, people just can just go 
csolve (x^3=27,x)... I wish I had had an 89 then (then I wouldn't have 
learned ANYTHING). Speaking of people who won't learn anything, someone in 
the ninth grade does NOT need an 89 (I know a 9th grader who has one). He's 
not going to learn anything...

Oh well, this degenerated into another rant about people using calcs too 
powerful for their needs...hehe =P

Michael

>
>Hi!
>
> > I _THINK_ that you'll only create a possible extraneous
> > root if either the numerator or denominator is even,
> > because that's the only time you have to worry about
> > positive and negative bases giving the same result when
> > raised to the power. In other words, if we had a^3=b^3,
> > I think we could make the assumption that a=b - but I'll
> > leave that to others here to confirm of refuse.
>
>Yes, this is for TI-Math, but I need some elaboration on
>it. This is true only if we assume the set of real numbers.
>a^n=b^n implies a=b if n is an odd number. But, in the set
>of complex numbers, this is not true. For example,
>
>a=2, b=-1+i*sqrt(3)
>
>Then a^3=b^3 (you can check it on TI), but a!=b...
>
>Now, the more general case, a^n=b^n. Assume that a is
>fixed. If n is integer, there is exactly n various
>possibilities for b (in the set of complex number) such
>that this equation is true. The situation is even worse
>if n is rational. And, if n is irrational, there exist
>an infinity number of different pairs (a,b) for which
>a^n=b^n, but the most two of them may be consist of
>real numbers.
>
>Cheers,
>
>Zeljko Juric
>
>P.S. If somebody has some advanced math question, he can
>try to ask me using my private mail: I am quite good in
>mathematics...
>

________________________________________________________________________
Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com