B Square root of a negative number in a complex field

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1. Nov 4, 2017

Staff: Mentor

To me the entire debate is about the inverse functions. $x \longmapsto x^2$ has none, since $x \longmapsto \pm \sqrt{x}$ is a relation, not a function. Similar happens with the exponential function. Whereas $\exp\, : \,(\mathbb{R,+}) \longrightarrow (\mathbb{R},\cdot)$ is an injective function which has a global inverse on its codomain, this is not true anymore for $\exp\, : \,(\mathbb{C,+}) \longrightarrow (\mathbb{C},\cdot)$. To make it injective one has to restrict its domain.

$\sqrt{a} = +\sqrt{a}$ for otherwise one has to write $\pm \sqrt{a}$.
$\exp(2n\pi i) = 1$ for all $n \in \mathbb{Z}$ which doesn't make the values $2n\pi$ equal to zero.
In my opinion the dispute is based on a wrong understanding of polar coordinates. Any equation $2n \pi i =0$ would immediately contradict the basic property $\operatorname{char} \mathbb{C} = 0$ and make the entire question meaningless as the numbers weren't defined anymore.

2. Nov 4, 2017

FactChecker

I'm sure the OP would not put it in those terms, but it is common to think of angles of 2πn equal to an angle of 0 and of rotations of 2πn to be the the same as the identity operation. In fact, it may even be the most common assumption. It can be done with mathematical rigor if necessary.

3. Nov 4, 2017

Staff: Mentor

But $U(1)$ and $\mathbb{C}$ are definitely not the same object, and I thought we were talking about complex numbers and not about the unitary group. At least the referenced article does.

4. Nov 5, 2017

WWGD

But the branch really lives in $U(1)/2n\pi^{-}$ all rotations except $2\pi$ or $0$, since $f(x)=f(x+2\pi)$ EDIT: Although this does not help much, since $U(1)-\{pt\} \neq \mathbb C$ ; it is actual homeo/iso to the Reals :(.

Last edited: Nov 5, 2017
5. Nov 5, 2017

Staff: Mentor

Or without removing this point $U(1) = \mathbb{P}(1,\mathbb{R})$.

6. Nov 5, 2017

WWGD

Aren't the inverses given in half-open(closed) strips, i.e., $a\leq \operatorname{im}(z) <a+2\pi$? ( we could, of course, also have the $\leq$ in the other end.)