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# help

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If $$x + \frac{1}{x} = 1$$, then find the value of $$x^{1000} + \frac{1}{x^{1000}}$$

Jul 6, 2020

#1
0

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

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

x = 1/2 + (i sqrt(3))/2 or x - 1/2 = -(i sqrt(3))/2

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

Jul 6, 2020
#2
+25541
+2

If  $$x + \dfrac{1}{x} = 1$$,  then find the value of  $$x^{1000} + \dfrac{1}{x^{1000}}$$

$$\text{Let x=e^{i\varphi} } \\ \text{Let \dfrac{1}{x}=x^{-1}=e^{-i\varphi}=e^{i(-\varphi)} } \\ \text{Let x^{1000}=e^{i(1000\varphi)} } \\ \text{Let \dfrac{1}{x^{1000}}=x^{-1000}=e^{-i\varphi 1000}=e^{i(-1000\varphi)} }$$

$$\begin{array}{|rcll|} \hline x + \dfrac{1}{x} &=& 1 \\ x + x^{-1} &=& 1 \\ e^{i\varphi} + e^{i(-\varphi)} &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(-\varphi)+i\sin(-\varphi) &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(\varphi)-i\sin(\varphi) &=& 1 \\ 2\cos(\varphi) &=& 1 \\ \cos(\varphi) &=& \dfrac{1}{2} \\ \varphi &=& \arccos\left( \dfrac{1}{2} \right) \\ \mathbf{ \varphi } &=& \mathbf{60^\circ} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \mathbf{x^{1000} + \dfrac{1}{x^{1000}}} \\ &=& e^{i(1000\varphi)} + e^{i(-1000\varphi)} \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(-1000\varphi)+i\sin(-1000\varphi) \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(1000\varphi)-i\sin(1000\varphi) \\ &=& 2\cos(1000\varphi) \quad | \quad \mathbf{ \varphi=60^\circ} \\ &=& 2\cos(1000*60^\circ) \\ &=& 2\cos(60000^\circ) \\ &=& 2\cos(240^\circ) \\ &=& 2(-0.5) \\ &=& \mathbf{-1} \\ \hline \end{array}$$

Jul 7, 2020
edited by heureka  Jul 7, 2020