# B Square root of a negative number in a complex field

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1. Nov 4, 2017

### FactChecker

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.

2. Nov 4, 2017

### Ivanovich62

Well, exactly.

Euler's formula is derived by defining a complex number as a position vector in 2d space. If you ignore that context, it will not make sense. However, if you do not ignore it, and realise that 2 pi i = 0 actually means unit vector with angle 2 pi = unit vector angle 0, it is correct. The mistake in C is using Euler's outside of that context.

3. Nov 4, 2017

### Staff: Mentor

$f(a)=f(b) \nRightarrow a=b$ - simple as that.

4. Nov 4, 2017

### Ivanovich62

unless f is 1-1

5. Nov 4, 2017

### Staff: Mentor

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

6. Nov 4, 2017

### Ivanovich62

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

7. Nov 4, 2017

### Staff: Mentor

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 $f(2\pi)$.

8. Nov 4, 2017

### Staff: Mentor

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.

9. Nov 4, 2017

### WWGD

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 $3=5$.

Last edited: Nov 4, 2017
10. Nov 4, 2017

### WWGD

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 $\sqrt{-100}*\sqrt{-100}=10i*10i=-100$

Last edited: Nov 4, 2017
11. Nov 4, 2017

### WWGD

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 $2n\pi$.

12. Nov 4, 2017

### Staff: Mentor

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.

13. Nov 4, 2017

### WWGD

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?

14. Nov 4, 2017

### Staff: Mentor

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.

15. Nov 4, 2017

### WWGD

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.

16. 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.

17. 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.

18. 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.

19. 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
20. Nov 5, 2017

### Staff: Mentor

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

21. 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.)