Sixth root of -64 using Euler's formula and De Moivre's theorem
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?
$\endgroup$ 21 Answer
$\begingroup$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.
$\endgroup$ 0More in general
"Zoraya ter Beek, age 29, just died by assisted suicide in the Netherlands. She was physically healthy, but psychologically depressed. It's an abomination that an entire society would actively facilitate, even encourage, someone ending their own life because they had no hope. Th…"
