Is $\sqrt[4]{-1}=\sqrt(i)$?

Question

I get $\sqrt[4]{-1}=((-1)^{\frac{1}{2}})^{\frac{1}{2}}=\sqrt{i}$.

But on the one hand $\sqrt[4]{-1}$ has $4$ roots and on the other $\sqrt{i}$ has just $2$.

Answer 1

Indeed as you observed $\sqrt[4]{-1}$ has four roots while $\sqrt i$ has only 2 roots thus they are not the same.

What is true is that they have roots in common.

Answer 2

For $k\in\mathbb{Z}$, we have $$-1=e^{i\pi}\implies (-1)^{1/4}=e^{i\pi(2k+1)/4}$$ so $$\sqrt[4]{-1}=e^{i\pi/4}, e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}$$ as $e^z$ has period $2i\pi$ with $z\in\mathbb{C}$.

whereas for $k\in\mathbb{Z}$, we have $$i=e^{i\pi/2}\implies i^{1/2}=e^{i\pi(k+1/4)}$$ so $$\sqrt i=e^{i\pi/4},e^{5\pi/4}$$ as $e^z$ has period $2i\pi$ with $z\in\mathbb{C}$.

Thus if you like, $\sqrt[4]{-1}$ contains $\sqrt i$.

Answer 3

The function $\sqrt[n]{x}$ is usually defined for non-negative, real $x$ and returns the non-negative, real solution to $y^n=x$. It can be extended to other domains, but must be accompanied by clear restrictions on which roots are to be returned.

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For example, we can define $\sqrt[4]{z}$ for $z\in\mathbb{C}$, where the root in the first quadrant (excluding the imaginary axis) is returned.

Similarly, we can define $\sqrt{z}$ for $z\in\mathbb{C}$, where the root in the first or second quadrant (excluding the negative real axis) is returned.

With these definitions, we get $\sqrt[4]{-1}=\sqrt{i\,}=\frac{1+i}{\sqrt2}$.

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However, if we consider $\sqrt[4]{z}$ to be the set valued function $\left\{w\in\mathbb{C}:w^4=z\right\}$, and similarly define $\sqrt{z}$ to be the set $\left\{w\in\mathbb{C}:w^2=z\right\}$, then $\sqrt[4]{-1}$ has $4$ elements and $\sqrt{i\,}$ has only two elements, so they are not the same, but with this definition, $$ \sqrt{i\,}\subset\sqrt[4]{-1} $$