B Square root of a negative number in a complex field

[Adgorn]

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?

 

[fresh_42]

It could be worth to read our insight article on the subject:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[Adgorn]

This helps, thanks :)

 

[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?

You may want to consider that $sqrt{100}=\pm 10$ and not necessarily 10. Similar for other roots. Depending on your choice of branch ( given that the root is defined in terms of the log) , you may have problems " jumping branches" .

 

[Mark44]

You may want to consider that $sqrt{100}=\pm 10$ and not necessarily 10.

Are you interpreting 100 as a complex number? If not, $\sqrt{100}$ is just plain old 10.

 

[WWGD]

Are you interpreting 100 as a complex number? If not, $\sqrt{100}$ is just plain old 10.

Yes, I sort of thought after I defined $\sqrt$ as I did, I would have to duck and cover. It is a pesky technical issue.

 

[Ivanovich62]

It could be worth to read our insight article on the subject:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

Example C states 2 pi i = 0, is this not true, since the 2 pi is an angle in radians

so the mistake is not considering it as an angle?

 

[Mark44]

Example C states 2 pi i = 0, is this not true, since the 2 pi is an angle in radians

so the mistake is not considering it as an angle?

No, it has nothing to do with being an angle. See the explanation for Example C a little way down the page.

 

[Ivanovich62]

That explanation depends on the 2 pi being an angle, and as far as I can remember, so does any proof of Euler's.

Is not the 2 pi the angle of the polar co-ordinate of a unit length position vector?

 

[fresh_42]

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.

This means that after a full circle, the function values are repeated and the inverse function cannot be taken anymore. So either we deal with infinitely many points all mapped to the same function value, or we restrict ourselves to angles strictly less than a full circle. Depending on the situation, both may be suitable to do. My teacher used to compare it with a spiral cut of a radish: Imagine you cut the complex number plane along the positive real axis. Then every full circle leads you to another next branch like in the picture here (just stacked):

Source: http://photos1.blogger.com/blogger/1879/2473/1600/daikon3upload.jpg

 

[FactChecker]

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

Not only is the square root of negative numbers defined in complex numbers, but many would consider that the main reason for using complex numbers.

The problem your friend points out can also be a problem in the square root of a positive number, since 10*10 = 100 and also -10*(-10) = 100. So it is just necessary to be careful with the square root function which is multiply defined.

 

[Ivanovich62]

yes, but this is because you are dealing with a 2d system rather than 1d, and all the error mentioned are due to considering the 2d system as 1d, when in fact the 1d system is just a special case of the 2d system, in particular, the fact that a complex number is a 2d position vector. the Euler equation only works taking this into account, that you are representing the vector in polar co-ordinates, but I could just as easily use degrees, or %age of full circle, or whatever angular measurement I choose to define. The point being: the fact that x is an angle is everything, not, as you say, nothing, and that is clearly the mistake being made in C. Since while 2 x 3.14... is not = 0, if we recognise that x is the angle of a unit vector polar co-ordinate in radians, as we should when using Euler's equation, then 2 x pi = 0 is correct.

 

[Mark44]

Since while 2 x 3.14... is not = 0, if we recognise that x is the angle of a unit vector polar co-ordinate in radians, as we should when using Euler's equation, then 2 x pi = 0 is correct.

Your argument doesn't hold water. An angle of $2\pi$ (or 360°) is not the same as an angle of 0 (or 0°), even though the positions of the terminal points of the rays that define each angle are equal. What you are saying is that there is no difference between one complete rotation and no rotations at all.

 

[Ivanovich62]

True, it is not the same angle, but the vectors (ie the complex numbers) with those angles are the same, that is my point.

 

[Mark44]

The problem your friend points out can also be a problem in the square root of a positive number, since 10*10 = 100 and also -10*(-10) = 100. So it is just necessary to be careful with the square root function which is multiply defined.

But by convention, the expression $\sqrt{100} = 10$, the principal square root of 100. For this reason it is erroneous to write $\sqrt{100} = \pm 10$.

 

[fresh_42]

True, it is not, but the vector (ie the complex number) is the same, that is my point.

In this case you placed a projection between the origin of your vector and the image after $n$ full circles. They are not the same thing, they are preimage and image. You must not arbitrarily identify them - there is a transformation in between!

 

[Ivanovich62]

so 2 x 1 = 2, this isn't the same 2, therefore 2 != 2 ?

 

[Mark44]

so 2 x 1 = 2, this isn't the same 2, therefore 2 != 2 ?

It is the same 2, since 1 is the multiplicative identity in the field of real number. IOW, if you multiply any real number by 1, you get back the same number.
But that's completely differernt from saying that $2\pi = 0$, which is obviously (I hope!) not true.

 

[FactChecker]

But by convention, the expression $\sqrt{100} = 10$, the principal square root of 100. For this reason it is erroneous to write $\sqrt{100} = \pm 10$.

That is true, but I assume that the OP did not intend that type of distinction in the context of his question. The logic that his friend was using did not recognize the principle branch or any convention for the √ symbol.

 

[Mark44]

That is true, but I assume that the OP did not intend that type of distinction in the context of his question. The logic that his friend was using did not recognize the principle branch or any convention for the √ symbol.

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.

You may want to consider that sqrt100=±10

The problem your friend points out can also be a problem in the square root of a positive number, since 10*10 = 100 and also -10*(-10) = 100. So it is just necessary to be careful with the square root function which is multiply defined.

... but not for positive real numbers.​

 

[Ivanovich62]

It is the same 2, since 1 is the multiplicative identity in the field of real number. IOW, if you multiply any real number by 1, you get back the same number.
But that's completely differernt from saying that $2\pi = 0$, which is obviously (I hope!) not true.

+ (whatever makes a circle) is additive identity for angle of polar position vectors

 

[Mark44]

+ (whatever makes a circle) is additive identity for angle of polar position vectors

You are confusing the concept of an angle with the terminal point for the angle. All of these angles are different, even though they all have the same terminal point:
$\{0, 2\pi, 4\pi, \dots, 2n\pi, \dots \}$

 

[Ivanovich62]

You are confusing the concept of an angle with the terminal point for the angle. All of these angles are different, even though they all have the same terminal point:
$\{0, 2\pi, 4\pi, \dots, 2n\pi, \dots \}$

no i am not, i agree they are different angles

however, the position vector of those angles is the same vector is the same point IS the same complex number.

 

[FactChecker]

I think we are giving the original post more thought than it is worth. The main point is that a person who works with roots of complex numbers must follow valid rules within that context to avoid errors like the OP.
@Adjorn is correct in his belief that the square root of negative numbers is an advantage of complex analysis. But it is multiply defined and must be done correctly.

 

[Mark44]

i agree they are different angles

Which makes the terminal points of the angles irrelevant. And the fact that they are different angles is the reason that the conclusion of Example C (in the Insights article quoted earlier) is a contradiction; i.e., that $2\pi i = 0$

 

[FactChecker]

Which makes the terminal points of the angles irrelevant. And the fact that they are different angles is the reason that the conclusion of Example C (in the Insights article quoted earlier) is a contradiction; i.e., that $2\pi i = 0$

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.

 

[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 realize 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.

 

[fresh_42]

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

 

[Ivanovich62]

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

unless f is 1-1

 

[Mark44]

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

unless f is 1-1

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

 

[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.)