Question

z^3=6 ( cos ( π/3 ) + i sin ( π/63 ) )

Answer 1

According to the De Moivre's Formula, all n-th roots of a complex number $r(\cos(\theta)+i\sin(\theta))$are given by $\sqrt[n]{r}\big(\cos({\theta+2\pi k\over n})+i\sin({\theta+2\pi k\over n})\big),$ $k=0,1,2,...,n-1$

We have that $r=6, \theta=\dfrac{\pi}{3}, n=3.$

$k=0:$

$$z_1=\sqrt[3]{6}\big(\cos({{\pi\over 3}+0\over 3})+i\sin({{\pi\over 3}+0\over 3})\big)=$$ $$=\sqrt[3]{6}\big(\cos({\pi\over 9})+i\sin({\pi\over 9})\big)=$$ $$=\sqrt[3]{6}\cos({\pi\over 9})+i\sqrt[3]{6}\sin({\pi\over 9})$$

$k=1:$

$$z_2=\sqrt[3]{6}\big(\cos({{\pi\over 3}+2\pi\over 3})+i\sin({{\pi\over 3}+2\pi\over 3})\big)=$$ $$=\sqrt[3]{6}\big(\cos({7\pi\over 9})+i\sin({7\pi\over 9})\big)=$$ $$=-\sqrt[3]{6}\cos({2\pi\over 9})+i\sqrt[3]{6}\sin({2\pi\over 9})$$

$k=2:$

$$z_3=\sqrt[3]{6}\big(\cos({{\pi\over 3}+4\pi\over 3})+i\sin({{\pi\over 3}+4\pi\over 3})\big)=$$ $$=\sqrt[3]{6}\big(\cos({13\pi\over 9})+i\sin({13\pi\over 9})\big)=$$ $$=-\sqrt[3]{6}\cos({4\pi\over 9})-i\sqrt[3]{6}\sin({4\pi\over 9})$$