If x+1/x=-1 then What is the value of x^247+1/x^187=?

May 28, 2016

$- 1$

Explanation:

$x = - \frac{1}{2} \pm i \frac{\sqrt{3}}{2} = \left(r \left(\cos \theta + i \sin \theta\right)\right) = r {e}^{i \theta} , r e s p e c t i v e l y$

r = 1.

For the first root, cosine is negative and sine is positive. $\theta$ is in

the second quadrant and is $\frac{2 \pi}{3}$.

For the second, both are negative. $\theta$ is in the 3rd quadrant.

So, it is $\frac{4 \pi}{3}$

Now, $x = {e}^{\frac{2 \pi}{3} i} \mathmr{and} {e}^{\frac{4 \pi}{3} i}$

Accordingly,

${x}^{247} + \frac{1}{x} ^ 247$

$= {e}^{247 \frac{2 \pi}{3} i} + {e}^{- 247 \frac{2 \pi}{3} i}$ and

${e}^{247 \frac{4 \pi}{3} i} + {e}^{- 247 \frac{4 \pi}{3} i}$

$= 2 \cos \left(247 \frac{2 \pi}{3} i\right) \mathmr{and} 2 \cos \left(247 \frac{4 \pi}{3} i\right)$

$= 2 \cos \left(164 \pi + \frac{2 \pi}{3}\right) \mathmr{and} 2 \cos \left(329 \pi + \frac{\pi}{3}\right)$

=2 cos (even $\pi + \frac{2 \pi}{3}$) and 2 cos (odd $\pi + \frac{\pi}{3}$)

$= 2 \cos \left(\frac{2 \pi}{3}\right) \mathmr{and} - 2 \cos \left(\frac{\pi}{3}\right)$

$= 2 \left(- \frac{1}{2}\right) \mathmr{and} \left(- 2\right) \left(\frac{1}{2}\right)$

Both are the same -1...

May 28, 2016

${x}^{247} + \frac{1}{{x}^{187}} = - 1$

Explanation:

In the equation $x + \frac{1}{x} = - 1$ occurs $\left\mid x \right\mid = 1$ because making $x = 1 + \delta$ and substituting there are not real solutions for $\delta$ in the equation $1 + \delta + \frac{1}{1 + \delta} = - 1$: then we can make $x = {e}^{i \theta}$.

Substituting we have

${e}^{i \theta} + {e}^{- i \theta} = - 1$

but

$\frac{{e}^{i \theta} + {e}^{- i \theta}}{2} = \cos \left(\theta\right)$

so the equation reduces to

$2 \cos \left(\theta\right) = - 1$. Solving for $\theta$ we have:

$\theta = \pm \frac{2}{3} \pi + 2 k \pi$ with $k = 0 , 1 , 2 , \ldots$

Now ${x}^{247} = {e}^{\pm i \times 247 \times \frac{2}{3} \pi} = {e}^{\pm i \frac{2}{3} \pi}$

and

${x}^{187} = {e}^{\pm i \times 187 \times \frac{2}{3} \pi} = {e}^{\pm i \frac{2}{3} \pi}$

Putting all together

${x}^{247} + \frac{1}{{x}^{187}} = 2 \cos \left(\pm \frac{2}{3} \pi\right) = 2 \left(- \frac{1}{2}\right) = - 1$

May 28, 2016

-1

Explanation:

For lower class student I do it as follows

Given

$x + \frac{1}{x} = - 1 \implies {x}^{2} + x + 1 = 0$

So

${x}^{3} = {x}^{3} - 1 + 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right) - 1 = \left(x - 1\right) \times 0 - 1 = - 1$

Now

${x}^{247} + \frac{1}{x} ^ 187 = {\left({x}^{3}\right)}^{82} \cdot x + \frac{1}{{\left({x}^{3}\right)}^{62} \cdot x} = \left({1}^{82} \cdot x\right) + \frac{1}{{1}^{62} \cdot x} = x + \frac{1}{x} = - 1$

Will this do?