The norm and the angle of the complex number $\sqrt[3]{-1}\sqrt[6]{7}$?

Question

I am learning the roots of unity here. I want to express arbitrary complex number in radical form in terms of the norm and the angle to use the formula $e^{i\phi}=\cos(\phi)+i\sin(\phi)$. Consider $x^6=7$ that is not of the roots of unity form such as $x^6=1$. Now its solution is $x=\sqrt[3]{-1}\sqrt[6]{7}\in\mathbb C$: how to calculate the norm and the angle for $\sqrt[3]{-1}\sqrt[6]{7}$?

Answer 1

Write $x$ in polar form: $$ x = re^{i\phi}. $$ Then the equation $x^6 = 7$ becomes $$ r^6 e^{6i\phi} = 7e^{0} $$ so $$ r^6 = 7 \qquad\Longrightarrow\qquad r = \sqrt[6]{7} $$ and $$ 6\phi \equiv 0 \pmod{2\pi} \qquad\Longrightarrow\qquad \phi \equiv 0 \pmod{\tfrac{\pi}{3}}. $$ The last condition is equivalent to being a cube root of $-1 = e^{i\pi}$.