B Square root of a negative number in a complex field

[Ivanovich62]

And clearly, the function we're talking about here is not 1-1.

it is if you restrict x to 0 - 2 pi radian, which you would do when defining a position vector as polar

 

[Mark44]

it is if you restrict x to 0 - 2 pi radian, which you would do when defining a position vector as polar

Still not 1-1, as there are two angles that map to the same point, if the interval you're talking about is $[0, 2\pi]$. If you're talking about the half-open interval $[0, 2\pi)$, then $2\pi$ isn't in that interval, so it doesn't make sense to compar $f(0)$ and $f(2\pi)$.

 

[Mark44]

To get us back from our off-topic foray, Example C in the linked Insights article arrived at the conclusion that $2\pi i = 0$. It ought to be obvious to anyone that this is patently false. In the complex plane, $2\pi i$ is more than 6 units up the vertical axis. Treating $2\pi$ and 0 as angles in radian measure, they are clearly different angles.

 

[WWGD]

I get that, but my comment had to do with a couple of posts that deal strictly with the square root of a positive real number.

... but not for positive real numbers.​

This is a sort of confusing , somewhat contentious issue as to how a square root is defined, but, I think the fact that there are two solutions does , I believe, contribute to the confusion in this situation. EDIT: Many , in order to define $\sqrt {x}$ as a function, select just the positive solution, and, yes, if we do not do this, then $\sqrt {x}$ is not a function in the standard sense, but instead a multi-function. Still, the problem is the assumption $\sqrt{ 10,000}= 100=-100i$ implies $100=-100i$ ; assuming there can only be one solution. It is just like assuming that if $f(3)=f(5)=0$ in $x^2-8x+15$ implies that $3=5$.

 

[WWGD]

Mod note: Fixed all of the radicals. The expressions inside the radical need to be surrounded with braces -- { }
(This question is probably asked a lot but I could not find it so I'll just ask it myself.)
Does the square root of negative numbers exist in the complex field? In other words is $\sqrt{(-a)}=\sqrt {(a)}*i$ or is it undefined? Recently in a forum I saw a guy claiming that the square root of -10 is undefined and his proof was something like this:

If $\sqrt{-100}=10i$ then $\sqrt{-100}*\sqrt{-100}=10i*10i=-100$, but
$\sqrt{-100}*\sqrt{-100}=\sqrt{-100*-100}=\sqrt{10,000}=\sqrt{100*100}=\sqrt{100}*\sqrt{100}=100$
Which implies $100=-100$, a paradox. And so $\sqrt{-100}$ is not $10i$ but is instead undefined.

I feel like he is making some sort of assumption here that I am not seeing because the square root of negatives is undefined in the complex field that kind of defeats the point of having the complex field in the first place, so is he right?

No, the problem is you are assuming the expression $\sqrt{x}$ can have just one solution. If you drop that assumption there is no paradox. You may have just one solution within a cut of width $\pi$. And, no, the square root of a negative number _is_ defined in the field of Complex numbers; you made use of this definition yourself when you wrote $\sqrt{-100}*\sqrt{-100}=10i*10i=-100$

 

[WWGD]

To get us back from our off-topic foray, Example C in the linked Insights article arrived at the conclusion that $2\pi i = 0$. It ought to be obvious to anyone that this is patently false. In the complex plane, $2\pi i$ is more than 6 units up the vertical axis. Treating $2\pi$ and 0 as angles in radian measure, they are clearly different angles.

But I think it ultimately depends on your frame of reference for angles, i.e.,what you set as your starting angle along the x-axis. You may set this angle to be 0 , which is then "equal" to $2n\pi$.

 

[Mark44]

This is a sort of confusing , somewhat contentious issue as to how a square root is defined, but, I think the fact that there are two solutions does , I believe, contribute to the confusion in this situation.

Really, the only confusion comes from not recognizing the difference between, say, the equation $x^2 - 10000 = 0$, which has two solutions, and the expression $\sqrt{10000}$, which represents a single value, 100.

EDIT: Many , in order to define $\sqrt {x}$ as a function, select just the positive solution, and, yes, if we do not do this, then $\sqrt {x}$ is not a function in the standard sense, but instead a multi-function.

Or a nonfunction, using the usual definition of a real function of a real variable.

Still, the problem is the assumption $\sqrt{ 10,000}= 100=-100i$ implies $100=-100i$ ; assuming there can only be one solution.

?
How does it make sense to start with the assumption that $\sqrt{ 10,000}= 100=-100i$? In particular, the part with 100 being equal to -100i?
If you start with an assumption that is false, the conclusion portion can be any statement, and the overall implication will be true, but meaningless.

 

[WWGD]

Really, the only confusion comes from not recognizing the difference between, say, the equation $x^2 - 10000 = 0$, which has two solutions, and the expression $\sqrt{10000}$, which represents a single value, 100.
Or a nonfunction, using the usual definition of a real function of a real variable.?
How does it make sense to start with the assumption that $\sqrt{ 10,000}= 100=-100i$? In particular, the part with 100 being equal to -100i?
If you start with an assumption that is false, the conclusion portion can be any statement, and the overall implication will be true, but meaningless.

No, I am not starting from the assumption; this is the conclusion and I am criticquing it. The rest is just a matter of semantics and choice of definitions, which differe from place-to-place. And I don't understanding just what you are driving at: yes, a function has a unique output. Have I said or implied otherwise anywhere?

 

[Mark44]

That depends on the context. To say that they are in the same equivalence class in certain contexts is to say that they are equal in those contexts.

This makes sense, but I don't believe that the OP asked the question in the context of equivalence classes.

But I think it ultimately depends on your frame of reference for angles, i.e.,what you set as your starting angle along the x-axis. You may set this angle to be 0 , which is then "equal" to $2n\pi$.

If we're talking equivalence classes, then yes, but otherwise $2\pi \ne 0$. In such cases it is sophistry to try to convince someone that the two are equal.

 

[WWGD]

This makes sense, but I don't believe that the OP asked the question in the context of equivalence classes.If we're talking equivalence classes, then yes, but otherwise $2\pi \ne 0$. In such cases it is sophistry to try to convince someone that the two are equal.

What I mean is if we choose to identify the positive Real axis with the angle 0 , then every loop around will coincide, as a position vector, with 0 , unless we do some Riemann surface or something of the sort. But, no, as standard Real numbers, I do _not_ state nor believe that $2\pi=0$. Maybe in some weird field or Mathematical object this may be true but, as standard Real numbers they are not equal.

 

[fresh_42]

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.

 

[FactChecker]

This makes sense, but I don't believe that the OP asked the question in the context of equivalence classes.

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.

 

[fresh_42]

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.

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.

 

[WWGD]

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.

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 :(.

 

[fresh_42]

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 :(.

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

 

[WWGD]

The point is, that the exponential (logarithm) function behaves differently as real or complex function.
The complex exponential function is periodic with the complex period $2\pi \mathrm{i}$, i.e. not bijective anymore. $\displaystyle \exp (z + 2 \pi k \mathrm{i}) = \exp (z), \; k \in \mathbb {Z}.$ By restriction of the domain to a strip $\displaystyle \{z \in \mathbb{C} \, | \, a <\operatorname{im} (z) < a + 2 \pi \}$ with $\displaystyle a \in \mathbb{R}$ it has a well-defined inverse, the complex logarithm.

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.)