Question 1207934: If x² - x + 1 = 0
then x²⁰²⁰ + x¹⁰¹⁰ - 1 is equal to
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52752)   (Show Source):
You can put this solution on YOUR website! .
If x^2 - x + 1 = 0, then x^2020 + x^1010 - 1 is equal to
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Let
 -  +  =  . (1)
Multiply both sides by (x+1). You will get
 + = , or =  .
+---------------------------------------------------------------+
| So, x is a complex number, which is a cubic root of -1. |
+---------------------------------------------------------------+
2020 = 673*3 + 1. Therefore
 =  =  = -x.
Next, 1010 = 336*3 + 2. Therefore
 =  =  = .
Thus, + =  + = - .
But, due to (1), - = .
Therefore, + = , or + + = 0.
Now subtract 2 from both sides of the last equality. You will get
+ - =  .
At this point, the solution is complete.
ANSWER. If - + = , then + - is equal to -2.
Solved.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Answer: -2
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Explanation
The other tutor has a great solution.
I'll show an alternative pathway.
Perhaps far less efficient but it's another way of thinking of this type of problem.
Use the quadratic formula to solve
x^2 - x + 1 = 0
and you'll get these complex roots
x = (1/2) + (sqrt(3)/2)*i
x = (1/2) - (sqrt(3)/2)*i
I skipped showing steps since many online calculators can provide them.
This solution is fairly lengthy as it is.
Let me know if you need to see the steps.
Convert each cartesian form to polar form.
You should get these two roots
x = cis(pi/3)
x = cis(-pi/3)
where "cis" is shorthand for "cosine i sine".
So for example,
cis(pi/3) = cos(pi/3) + i*sin(pi/3)
Why are we dealing with polar form?
To take advantage of De Moivre's Theorem.
That theorem is this:
if z = r*cis(theta) then z^n = r^n*cis(n*theta) where n is a positive integer.
Because sine and cosine have period 2pi radians, we have this very useful property
cis(x + 2pi*n) = cis(x)
where n is any integer.
Adding on one copy of 2pi will rotate the angle a full revolution.
But the angle remains pointing in the same direction (which explains why sine & cosine don't change).
The same will apply when adding on multiple copies of 2pi radians.
Let's use De Moivre's Theorem to compute x^1010 for the first root mentioned.
x = cis(pi/3)
x^1010 = cis(1010pi/3)
x^1010 = cis(2pi/3 + 1008pi/3)
x^1010 = cis(2pi/3 + 336pi)
x^1010 = cis(2pi/3 + 2pi*168)
x^1010 = cis(2pi/3) ..................... use the formula mentioned above
x^1010 = cos(2pi/3) + i*sin(2pi/3)
x^1010 = (-1/2) + (sqrt(3)/2)*i ............... Use the unit circle
Use WolframAlpha, GeoGebra, or similar to verify this portion is correct so far.
Another application of De Moivre's Theorem can be done to find x^2020, or you can square the value of x^1010 to avoid De Moivre.
Here's what it looks like when using De Moivre's Theorem.
x = cis(pi/3)
x^2020 = cis(2020*pi/3)
x^2020 = cis(4pi/3+2016pi/3)
x^2020 = cis(4pi/3+672pi)
x^2020 = cis(4pi/3+2pi*336)
x^2020 = cis(4pi/3) ................ use the formula mentioned above
x^2020 = cos(4pi/3) + i*sin(4pi/3)
x^2020 = (-1/2) - (sqrt(3)/2)*i
OR
If you want to avoid De Moivre, then,
x^1010 = (-1/2) + (sqrt(3)/2)*i
(x^1010)^2 = [ (-1/2) + (sqrt(3)/2)*i ]^2
x^2020 = (-1/2)^2 + 2*(-1/2)*(i*sqrt(3)/2) + (i*sqrt(3)/2)^2
x^2020 = 1/4 - (sqrt(3)/2)*i - 3/4
x^2020 = (-1/2) - (sqrt(3)/2)*i
I used the formula (a+b)^2 = a^2+2ab+b^2 on the third step.
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If x = cis(pi/3) then it leads to these values
x^2020 = (-1/2) - (sqrt(3)/2)*i
x^1010 = (-1/2) + (sqrt(3)/2)*i
Add straight down to arrive at the sum -1. The imaginary parts cancel out.
Therefore, x^2020+x^1010 = -1 when x = cis(pi/3).
Subtract 1 from both sides to get x^2020+x^1010-1 = -2
You should find that x = cis(-pi/3) leads to the same answer.
A similar question is found here
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