Sixth root of -64 using Euler's formula and De Moivre's theorem

Question

I am attempting to solve: $$(-64)^{\frac{1}{6}}$$ Using the relation: $$a+bi=re^{i(\tan^{-1}(\frac{b}{a})+2\pi n)}$$ And then applying De Moivre's theorem: $$(a+bi)^{\frac{1}{x}}=(re^{i(\tan^{-1}(\frac{b}{a})+2\pi n)})^{\frac{1}{x}}$$ $$...=r^{\frac{1}{x}}e^{i(\frac{\tan^{-1}(\frac{b}{a})}{x}+\frac{2\pi n}{x})}$$ $$...=64^{\frac{1}{6}}e^{i(\frac{\tan^{-1}(\frac{0}{64})}{6}+\frac{2\pi n}{6})}$$ $$...=2e^{i(\frac{\pi n}{3})}$$ Now! When $n=0$ (the first case in using this strategy), then we get $$2e^{0i}$$ Which should be, drumroll, $2$. Obviously though, $2$ is not a sixth root of $-64$. So where did this methodology go wrong?

Answer 1

Your problem arises from assuming that the principal argument of a complex number $\text{a + ib}$ is equal to $\arctan{\frac{b}{a}}$.

This would assume your argument to be 0, when in fact it is equal to $\pi$.

Try your calculations again using this.