Radical problem: $\sqrt[6]{-1}=\sqrt{\sqrt[3]{-1}}=\sqrt {-1}=i$

Question

I'm stuck in a very simple problem . But, I could not find an answer in the books. That's why I want to ask MSE.

Why is not this correct (according to Wolfram Alpha) ?

$\sqrt[6]{-1}=\sqrt{\sqrt[3]{-1}}=\sqrt {-1}=i$

$\sqrt[3]{-1}=-1$, Because, $(-1)^3=-1.$

Where is the mistake?

Answer 1

It's "correct" in the sense that $i^6=(-1)^3=-1$, but many solutions are missed. The solutions of $z^3=-1$ are $-e^{2\pi ik/3}$ with $k\in\{0,\,1,\,2\}$, while those of $z^6=-1$ are $ie^{\pi ik/3}$ with $k\in\{0,\,1,\,2,\,3,\,4,\,5\}$.

Answer 2

$-1$ has six different sixth roots. You listed one of them; there are still five more.

To write $\sqrt[6]{-1}$ as if it is a well-defined notation requires a convention to decide which of the sixth roots to give. The usual convention is not the one you listed.

Note, incidentally, that the usual convention also has $\sqrt[3]{-1} \neq -1$. When moving away from the real numbers to the more general setting of the complex numbers, we no longer use the convention based on the weird property of real numbers that every real number has a unique cube root.