SVP tres important!!

Bonjour  Reinaflayfel

[tex](\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=(e^{i\frac{\pi}{6}})^{1999}\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=(e^{i\dfrac{1999\pi}{6}})\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=\cos(\dfrac{1999\pi}{6})+i\sin(\dfrac{1999\pi}{6})\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=\cos(\dfrac{2004\pi}{6}-\dfrac{5\pi}{6})+i\sin(\dfrac{2004\pi}{6}-\dfrac{5\pi}{6})\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=\cos(334\pi-\dfrac{5\pi}{6})+i\sin(334\pi-\dfrac{5\pi}{6})[/tex]

[tex]\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=\cos(137\times2\pi-\dfrac{5\pi}{6})+i\sin(137\times2\pi-\dfrac{5\pi}{6})\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=\cos(-\dfrac{5\pi}{6})+i\sin(-\dfrac{5\pi}{6})\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=\cos(\dfrac{5\pi}{6})-i\sin(\dfrac{5\pi}{6})\\\\(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=-\cos(\dfrac{\pi}{6})-i\sin(\dfrac{\pi}{6})\\\\\boxed{(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i)^{1999}=-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}}[/tex]