Given that (a+bi)^2002=a-bi, how many solutions are there for (a, b)?
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Given that (a+bi)^2002=a-bi, how many solutions are there for (a, b)?
(r,t)^2002=(r,-t)=(r^2002,2002t)
that gives:
r^2002=r, 2002t=-t+2*pi*k
r^2002-r=0 yields r=0 or r=1, since r is real.
r=0 yields a single solution.
r=1 yields:
2003t=2*pi*k
since t is between 0 and 2*pi, then we are confident that 2003pi*k>2003t>=0
ie k=0 to 2002, yielding 2003 solutions. plus 1 from before to get 2004
What is the value of i? :(
i2=-1
-|: o)
:rolleyes:
OH i, as in Imaginary
i^2=-1
sqr(i) = -1
NO! SQRT(i) is not -1!! only i^2=-1!!!!!!!!!!!!!!!!!!
I think your analysis has a problem.
See:
http://monet.physik.unibas.ch/~elmer...m/polyroot.htm
my analysis? whats the problem (i found nothing in that site)
The thing is, if you'd told us that question was from a recent math challenge, the answer almost HAD to be "2003" didn't it. If not, that's a mighty big coincidence.
its a multiple choice question from 2002 American Invitational Challenge (let me mention that as you know we are from Windsor Ontario, which is NOT part of the states but rather Canada ;))
btw the choices were (if my memories serve me correct)
a) 2001 b) 2002 c) 2003 d) 2004 e) 2005
Well, at least you didn't say i = (-1)(1/2)! Bugz would have had a heart attack!Quote:
Originally posted by alkatran
:rolleyes:
sqr(i) = -1
:p
But, just to clarify on what you did say,
In actuality,
sqr(i) = (1 + i)/sqr(2)
{Or it could also be expressed as: sqr(i) = Pos or Neg((sqr(2)/2)*(1+i))}
;)
-Lou