Complex Numbers

[Kalkewl8ter]

Given that (a+bi)^2002=a-bi, how many solutions are there for (a, b)?

[bugzpodder]

(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

[alkatran]

What is the value of i?

[bugzpodder]

i²=-1

-|: o)

[alkatran]

OH i, as in Imaginary
i^2=-1
sqr(i) = -1

[bugzpodder]

NO! SQRT(i) is not -1!! only i^2=-1!!!!!!!!!!!!!!!!!!

[jim mcnamara]

I think your analysis has a problem.

See:

http://monet.physik.unibas.ch/~elmer...m/polyroot.htm

[bugzpodder]

my analysis? whats the problem (i found nothing in that site)

[A$$Bandit]

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.

[bugzpodder]

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

[NotLKH]

Originally posted by alkatran

sqr(i) = -1

Well, at least you didn't say i = (-1)^(1/2)! Bugz would have had a heart attack!


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