trigonometric form and DeMoivres Theorem

**kenzie103109** · May 19th 2010, 08:14 AM

list the 3 complex roots of -1 by rewriting -1 in trogonometric form and applying DeMoivre's Theorem. you may leave your answers in trigonometris form

**tonio** · May 19th 2010, 11:29 AM

Originally Posted by kenzie103109

list the 3 complex roots of -1 by rewriting -1 in trogonometric form and applying DeMoivre's Theorem. you may leave your answers in trigonometris form

$z^3=-1=cis(\pi+2k\pi)=cis(\pi(2k+1))\,,\,\,with\,\,\,ci s(x):=e^{xi}\,,\,\,x\in\mathbb{R}\,,\,\,k\in\mathb b{Z}$ . In order to get all the different elements though we restrict $k=0,1,2$ (why? Check the trigonometric circle).

So $z=(-1)^{1/3}=cis(\pi(2k+1))^{1/3}=cis\left(\pi\frac{2k+1}{3}\right)$ (here we used De Moivre's theorem) , so now just input $k=0,1,2$ and obtain the three complex roots of $-1$ ...as simple as that!

Tonio

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