Đặt $z=\cos\varphi+i\sin\varphi\implies z+\frac 1z=2\cos\varphi=1$
Vì $\cos $ đối và có chu kỳ $2\pi$ nên ta chọn $\varphi=\frac\pi3$.
Ta có: $z^n+\frac1{z^n}=2\cos \frac{n\pi}{3}$ do đó:
$S=2\left(\cos\frac\pi3+\cdots+\cos\frac{n\pi}{3}\right)\\\iff\frac{\sqrt3}2\cdot S=\sin\frac\pi3\cdot 2\left(\cos\frac\pi3+\cdots+\cos\frac{n\pi}{3}\right) \\\iff \frac{\sqrt3}2\cdot S = \sin\frac{2\pi}3+\sin\frac{3\pi}{3}-\sin\frac{\pi}{3}+\sin\frac{4\pi}{3}-\sin\frac{2\pi}3+\cdots+\sin\frac{(n+1)\pi}{3}-\sin\frac{(n-1)\pi}{3}\\ \iff \frac{\sqrt3}2\cdot S = -\sin\frac{\pi}{3}+ \sin\frac{n\pi}{3}+\sin\frac{(n+1)\pi}{3}\\ \iff \frac{\sqrt3}2\cdot S = -\frac{\sqrt3}2+\sqrt3\sin\frac{(2n+1)\pi}{6}\\ \iff S=-1+2 \sin\frac{(2n+1)\pi}{6}$
Vẽ đường tròn lượng giác, ta có:
- $n=6k, (k\in \mathbb{N}, n\neq 0)\implies S=-1+2\sin\frac{(12k+1)\pi}{6}=-1+2\sin\frac{\pi}{6}=0$
- $n=6k+1\implies S=-1+2 \sin\frac{(12k+3)\pi}{6}=1$
- $n=6k+2\implies S=-1+2\sin\frac{(12k+5)\pi}{6}=0$
- $n=6k+3\implies S=-1+2\sin\frac{(12k+7)\pi}6=-2$
- $n=6k+4\implies S=-1+2\sin\frac{(12k+9)\pi}6=-3$
- $n=6k+5\implies S=-1+2\sin\frac{(12k+11)\pi}6=-2.\blacksquare$
LZuTao muốn đặt z$z = cos\varphi +i.sin\varphi$ thì phải chi ra |z| = 1