Question

Using De Moivre's theorem or otherwise
find the six roots of the equation z^6 +1 = 0
giving your answers in the form e^iw where w= angle teta

Answer 1

Let us find the six roots of the equation $z^6=-1$ using de Moivre's Theorem for fractional power:

$[r(\cos\theta+i\sin\theta)]^{\frac{1}{n}}=\sqrt[n]{r}(\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n})$ for $k\in\{0,1,...,n-1\}.$

In trigonometric form $-1=\cos\pi+i\sin\pi$, that is $r=1, \theta=\pi.$ Consequently, the formula

$z_k=\cos\frac{\pi+2\pi k}{6}+i\sin\frac{\pi+2\pi k}{6}=e^{\frac{\pi+2\pi k}{6}i}$ for $k\in\{0,1,2,3,4,5\}$ gives all six different roots.

Therefore, we have:

$z_0=e^{\frac{\pi}{6}i}$, $z_1=e^{\frac{3\pi}{6}i}=e^{\frac{\pi}{2}i}$, $z_2=e^{\frac{5\pi}{6}i}$, $z_3=e^{\frac{7\pi}{6}i}$, $z_4=e^{\frac{9\pi}{6}i}=e^{\frac{3\pi}{2}i}$, $z_5=e^{\frac{11\pi}{6}i}.$