Question

3. Use De Moivre’s Theorem to determine the cube root of Z and leave your answer in polar

form with the angle in radians

(a) Z = 1+i√3

Answer 1

At first, we rewrite $z=1+i\sqrt{3}$ in the polar form. We get: $z=2(\frac12+i\frac{\sqrt{3}}2)=2(\cos(\frac{\pi}{3})+i\,\sin(\frac{\pi}{3}))$.

We use the known formula (extension of de Moivre's formula) to compute the root $\frac13$. We get:

$z^{\frac13}=2^{\frac13}(\cos(\frac{\frac{\pi}3+2\pi k}{3})+i\,\sin(\frac{\frac{\pi}3+2\pi k}{3})),$ where $k=0,1,2.$

We receive the following roots:

$v_0=2^{\frac13}(\cos(\frac{{\pi}}{9})+i\,\sin(\frac{{\pi}}{9}))$,

$v_1=2^{\frac13}(\cos(\frac{{7\pi}}{9})+i\,\sin(\frac{{7\pi}}{9})),$

$v_2=2^{\frac13}(-\cos(\frac{{4\pi}}{9})-i\,\sin(\frac{{4\pi}}{9}))$.

Answer: the cube root of $z$ has the following values:

$v_0=2^{\frac13}(\cos(\frac{{\pi}}{9})+i\,\sin(\frac{{\pi}}{9}))$,

$v_1=2^{\frac13}(\cos(\frac{{7\pi}}{9})+i\,\sin(\frac{{7\pi}}{9})),$

$v_2=2^{\frac13}(-\cos(\frac{{4\pi}}{9})-i\,\sin(\frac{{4\pi}}{9}))$.