\(\)If \(x + \frac{1}{x} = 1\), then find the value of \(x^{1000} + \frac{1}{x^{1000}}\)
Solve for x:
x + 1/x = 1
Bring x + 1/x together using the common denominator x:
(x^2 + 1)/x = 1
Multiply both sides by x:
x^2 + 1 = x
Subtract x from both sides:
x^2 - x + 1 = 0
Subtract 1 from both sides:
x^2 - x = -1
Add 1/4 to both sides:
x^2 - x + 1/4 = -3/4
Write the left hand side as a square:
(x - 1/2)^2 = -3/4
Take the square root of both sides:
x - 1/2 = (i sqrt(3))/2 or x - 1/2 = -(i sqrt(3))/2
Add 1/2 to both sides:
x = 1/2 + (i sqrt(3))/2 or x - 1/2 = -(i sqrt(3))/2
Add 1/2 to both sides:
x = 1/2 + (i sqrt(3))/2 or x = 1/2 - (i sqrt(3))/2
[1/2 + (i sqrt(3))/2]^1000 + 1/ [1/2 + (i sqrt(3))/2]^1000 = - 1
+26367 If \(x + \dfrac{1}{x} = 1\), then find the value of \(x^{1000} + \dfrac{1}{x^{1000}}\)
\(\text{Let $x=e^{i\varphi} $} \\ \text{Let $\dfrac{1}{x}=x^{-1}=e^{-i\varphi}=e^{i(-\varphi)} $} \\ \text{Let $x^{1000}=e^{i(1000\varphi)} $} \\ \text{Let $\dfrac{1}{x^{1000}}=x^{-1000}=e^{-i\varphi 1000}=e^{i(-1000\varphi)} $}\)
\(\begin{array}{|rcll|} \hline x + \dfrac{1}{x} &=& 1 \\ x + x^{-1} &=& 1 \\ e^{i\varphi} + e^{i(-\varphi)} &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(-\varphi)+i\sin(-\varphi) &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(\varphi)-i\sin(\varphi) &=& 1 \\ 2\cos(\varphi) &=& 1 \\ \cos(\varphi) &=& \dfrac{1}{2} \\ \varphi &=& \arccos\left( \dfrac{1}{2} \right) \\ \mathbf{ \varphi } &=& \mathbf{60^\circ} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline && \mathbf{x^{1000} + \dfrac{1}{x^{1000}}} \\ &=& e^{i(1000\varphi)} + e^{i(-1000\varphi)} \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(-1000\varphi)+i\sin(-1000\varphi) \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(1000\varphi)-i\sin(1000\varphi) \\ &=& 2\cos(1000\varphi) \quad | \quad \mathbf{ \varphi=60^\circ} \\ &=& 2\cos(1000*60^\circ) \\ &=& 2\cos(60000^\circ) \\ &=& 2\cos(240^\circ) \\ &=& 2(-0.5) \\ &=& \mathbf{-1} \\ \hline \end{array}\)
