1. ## Little Question

if $x+\frac{1}{x}=-1$.
What's the value of $x^{2007}+\frac{1}{x^{2007}}$?

I try using this:

$x^{2007}+\frac{1}{x^{2007}}=\left(x+\frac{1}{x}\ri ght)^{2007}-\sum_{k=1}^{2006}\binom{2007}{k}x^{2k-2007}$

but i don't find any results.

2. On the basis of the identity...

$x + \frac{1}{x} = e^{\ln x} + e^{-\ln x}= 2\cdot \cosh(\ln x) = 2\cdot \cos (i\cdot \ln x)$ (1)

... we can write...

$x + \frac{1}{x} = -1 \rightarrow \ln x = i\cdot \frac{4}{3}\cdot \pi$ (2)

Since 2007 devides 3 is $\ln x^{2007} = 2007\cdot \ln x = 4\cdot i\cdot k\cdot \pi$ so that is...

$x^{2007} + x^{-2007} = 2\cdot \cos (4\cdot k \cdot \pi)= 2$ (3)

Kind regards

$\chi$ $\sigma$

3. Originally Posted by streethot
if $x+\frac{1}{x}=-1$.
What's the value of $x^{2007}+\frac{1}{x^{2007}}$?

I try using this:

$x^{2007}+\frac{1}{x^{2007}}=\left(x+\frac{1}{x}\ri ght)^{2007}-\sum_{k=1}^{2006}\binom{2007}{k}x^{2k-2007}$

but i don't find any results.
If:

$x+1/x=-1$

then $x=-1/2\pm \sqrt{3}/2=e^{2\pi i/3}\text{ or }e^{4\pi i/3}$

Then as $2007=3\times 669$ we have $x^{2007}=e^{1338 \pi i} \text{ or }e^{2676 \pi i} =1$ and so:

$
x^{2007}+x^{-2007}=1+1=2
$

cb

4. Then as we have and so:
$(e^{\frac{2i\pi}{3}})^{2007}=e^{i1338\pi}$

and

$(e^{\frac{4i\pi}{3}})^{2007}=e^{i2676\pi}$