Expert: Ahmed Salami - 5/27/2004

Question
Hello!
Can you help me with this problem:
If x1 and x2 are the solutions to this equation:x^2-x+1=0,  then x1^2004+x2^2004=?

Answer
Hi,
Sorry for the delay.
I hope you are familiar with complex numbers and their polar and exponential forms because you are going to need it.
For x^2 - x + 1 = 0
x = |1 +or- sqrt(1-4)|/2
x1 = |1 + sqrt(-3)|/2  =  |1 + i(sqrt3)|/2
x2 = |1 - sqrt(-3)|/2  =  |1 - i(sqrt3)|/2
where i = sqrt (-1)
Converting to the polar form,
z = r(cos# + isin#)
where r = sqrt (a^2 + b^2)
and  # = arctan (b/a)
Now x1 = (1/2)+ i(sqrt3)/2
r1 = sqrt|(1/2)^2 + (sqrt3/2)^2| = sqrt |(1/4)+(3/4)|
  = sqrt 1 = 1
#1 = arctan (sqrt3/2)/(1/2) = arctan (sqrt3) = 60
x2 = (1/2)- i(sqrt3)/2
r2 = sqrt|(1/2)^2 + (sqrt3/2)^2| = sqrt |(1/4)+(3/4)|
  = sqrt 1 = 1
#1 = arctan (-sqrt3/2)/(1/2) = arctan (-sqrt3) = -60 = 300
Therefore,
x1 = cos60 + isin60
x2 = cos300 + isin300
By De Moivre's theorem,
z^n = r^n(cosn# + isinn#)
Now,
x1^2004 = cos(2004)(60) + isin(2004)(60)
       = cos (120240) + isin (120240)
       = 1
x2^2004 = cos(2004)(300) + isin(2004)(300)
       = cos (601200) + isin (601200)
       = 1
we therefore have,
x1^2004 + x2^2004 = 1 + 1 = 2

Alternative method;
converting to exponential form,
z = re^i#
where e is the natural logarithm base and # is angle in radians
with r and # values as before
x1 = e^(pi/3)i
x2 = e^(5pi/3)i
therefore,
x1^2004 = |e^(pi/3)i|^2004 = e^(668pi)i
but the angle 668pi in radians is equivalent to 0 since it is just 334 revolutions plus 0.
x1^2004 = e^0i = e^0 = 1
similarly
x2^2004 = |e^(5pi/3)i|^2004 = e^(3340pi)i
also the angle 3340pi in radians is equivalent to 0 since it is just 1670 revolutions plus 0.
x2^2004 = e^0i = e^0 = 1
we therefore have,
x1^2004 + x2^2004 = 1 + 1 = 2
as before.

Hope it is helpful.
You can always ask more.