Question

If $z+\frac{1}{z}=2\cos 6^{\circ }$, then $z^{1000}+\frac{1}{z^{1000}}+1$ is equal to

Answer 1

Let $z=r(\cos \theta +i\sin \theta )$
$\Rightarrow \frac{1}{z}=\frac{1}{r}(\cos \theta -i\sin \theta )$
$\Rightarrow z+\frac{1}{z}=(r+\frac{1}{r})\cos \theta +i(r-\frac{1}{r})\sin \theta ...\left(1\right)$
$z+\frac{1}{z}=2\cos 6^{\circ }...\left(2\right)$
Equating $\left(1\right)$ and $\left(2\right)$, we get
$(r+\frac{1}{r})\cos \theta =2\cos 6^{\circ }$
$\theta =6^{\circ }$ and $r+\frac{1}{r}=2$
$\therefore r=1$

$\therefore z^{1000}+\frac{1}{z^{1000}}+1$
$=\cos {6000}^{\circ }+i\sin {6000}^{\circ }+\cos {6000}^{\circ }-i\sin {6000}^{\circ }+1$
$=2\cos {6000}^{\circ }+1$
$=2\cos {240}^{\circ }+1$
$=2\left(\frac{-1}{2}\right)+1=0.$