# If z+1/z=1, find z^3, and then z^1000+1/z^1000?

Aug 9, 2018

#### Explanation:

Let

$z = \left(\cos \theta + i \sin \theta\right)$

$\frac{1}{z} = \left(\cos \theta - i \sin \theta\right)$

$z + \frac{1}{z} = 2 \cos \theta = 1$

$\cos \theta = \frac{1}{2}$

$\theta = \frac{\pi}{3}$, $\left[\mod 2 \pi\right]$

Therefore,

$z = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)$

And

${z}^{3} = \cos \left(3 \cdot \frac{\pi}{3}\right) + i \sin \left(3 \cdot \frac{\pi}{3}\right) = - 1$

Avcoording to Demoivre' s Theorem

${z}^{1000} = \cos \left(1000 \cdot \frac{\pi}{3}\right) + i \sin \left(1000 \cdot \frac{\pi}{3}\right)$

$\frac{1}{z} ^ 1000 = \cos \left(1000 \cdot \frac{\pi}{3}\right) - i \sin \left(1000 \cdot \frac{\pi}{3}\right)$

${z}^{1000} + \frac{1}{z} ^ 1000 = 2 \cos \left(\frac{1000}{3} \pi\right) = - 2 \cdot 0.5 = - 1$

Aug 9, 2018

${z}^{3} = - 1 \text{ }$ and $\text{ } {z}^{1000} + \frac{1}{z} ^ 1000 = - 1$

#### Explanation:

Given:

$z + \frac{1}{z} = 1$

Note that:

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

So:

${z}^{3} = - 1$

So:

${z}^{1000} + \frac{1}{z} ^ 1000 = {z}^{3 \cdot 333 + 1} + \frac{1}{z} ^ \left(3 \cdot 333 + 1\right)$

$\textcolor{w h i t e}{{z}^{1000} + \frac{1}{z} ^ 1000} = {\left(- 1\right)}^{333} \left(z + \frac{1}{z}\right)$

$\textcolor{w h i t e}{{z}^{1000} + \frac{1}{z} ^ 1000} = - \left(z + \frac{1}{z}\right)$

$\textcolor{w h i t e}{{z}^{1000} + \frac{1}{z} ^ 1000} = - 1$