Question

In Exercises 81–86, solve each equation in the complex number system. Express solutions in polar and rectangular form. x⁶ − 1 = 0

Accepted Answer

Recognize that the equation $$x^{6} - 1 = 0$ can be rewritten as x$$^{6}$$ = 1. $$This means we are looking for the sixth roots of unity in the complex plane. Express the number 1 in polar form. Since 1 is on the positive real axis, its polar form is $$1 = 1 \left( \cos 0 + i \sin 0 ight)$$, or equivalently $$1 = 1 \operatorname{cis} 0$$ where $$\operatorname{cis} 	heta = \cos 	heta + i \sin 	heta. $$Use De Moivre's Theorem to find all sixth roots of unity. The general formula for the $$nth $$roots of a complex number $$r \operatorname{cis} 	heta is$$ $$r^{1/n}$$ \operatorname{cis} \left( \frac{	heta + 2k\pi}{n} ight)$$ for k = 0$$, 1, 2, ..., n-1$$. Here, r=1$, $$	heta=0$$, and n=6$. Calculate each root by substituting k = 0$$, 1, 2, 3, 4, 5$$ into the formula x_k = 1$$ \operatorname{cis} \left( \frac{2k\pi}{6} ight) = \cos \left( \frac{2k\pi}{6} ight) + i \sin \left( \frac{2k\pi}{6} ight)$$ to get all six solutions in polar and rectangular form. Write each solution explicitly in rectangular form as x_k$$ = \cos \left( \frac{2k\pi}{6} ight) + i \sin \left( \frac{2k\pi}{6} ight)$$ and in polar form as x_k = 1$$ \operatorname{cis} \left( \frac{2k\pi}{6} ight)$$ for k = 0$$, 1, 2, 3, 4, 5$$.

In Exercises 81–86, solve each equation in the complex number system. Express solutions in polar and rectangular form. x⁶ − 1 = 0

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Concetti chiave

Complex Roots of Unity

Polar Form of Complex Numbers

Conversion Between Polar and Rectangular Forms