How Are Square Roots Defined for Complex Numbers?

[Bacle2]

Mod note: These posts are orginally from the thread: https://www.physicsforums.com/showthread.php?t=626545


The square root is not defined everywhere, at least not as a function,

but as a multifunction, since every complex number has two square roots. I mean, the

expression z^1/2 is ambiguous until you choose a branch.

Sorry if you already are taking this into account; I am in nitpicking mode, but I

shouldn't take it out on you :) .

 

[chiro]

Don't mean to nitpick, but remember that it is for z in ℂ\{0} to start with; some profs.

may take away points in an exam if you don't specify this.

But also, remember your square root is not defined everywhere, at least not as a function,

but as a multifunction, since every complex number has two square roots. I mean, the

expression z^1/2 is ambiguous until you choose a branch.

Sorry if you already are taking this into account; I am in nitpicking mode, but I

shouldn't take it out on you :) .

Taking the square root of a number (even a complex one) is a function and is not multi-valued. I think you are confusing finding the roots of some expression with this and they are different.

Also it is true that any complex number can be written in infinitely many ways (by adding 2npi to the argument where n is an integer), but this is not the same as the issues with finding roots since the square root is a function and only returns one result (all functions have to do this). In terms of the principal argument, the result is always unique.

Your condition about not including 0 means that you can reverse r^1/2 to get r back and since this operation is a bijection, getting the inverse back with respect to the principle argument is also a bijection (and remember that r > 0 so we don't have any branch issues).

 

[Bacle2]

Taking the square root of a number (even a complex one) is a function and is not multi-valued. I think you are confusing finding the roots of some expression with this and they are different.

Also it is true that any complex number can be written in infinitely many ways (by adding 2npi to the argument where n is an integer), but this is not the same as the issues with finding roots since the square root is a function and only returns one result (all functions have to do this). In terms of the principal argument, the result is always unique.

Your condition about not including 0 means that you can reverse r^1/2 to get r back and since this operation is a bijection, getting the inverse back with respect to the principle argument is also a bijection (and remember that r > 0 so we don't have any branch issues).

What do you mean square roots are not multifunctions? They are multifunctions even in

the real case. In the complex case:

The square root of re^iθ are:

r^1/2e^(iθ/2) and

r^1/2e^i(θ/2+∏) , by.e.g. DeMoivre's inequality:

and the roots are ∏<2∏ appart, so they are not separated by different copies of the plane

as, e.g., periodic functions of period ≥2∏.

 

[pwsnafu]

They are multifunctions even in

the real case.

The square root of a non-negative real number is always non-negative. It is never multi-valued.

 

[Bacle2]

Well, I think this is a technical/language issue: by default, one often _defines_ the

square root √x is defined by default to be the positive solution--when it exists, i.e.,

when x>0 ( for 0, there is one solution). Maybe a more accurate way for me to

phrase things is : when x is a positive number, there are two numbers whose square

equals x. I don't know if the choice of the positive solution is just a convention, or if

there is some other reason for this choice.

 

[chiro]

Finding roots is not the same as finding the function to some power: they are completely different things.

Again, functions always by definition return one output: they have to.

Root-finding is an entirely different thing and not to be confused with an actual function.

 

[Bacle2]

Yes, that is the precise point I am making, and this is why I'm making the distinction

between multifunctions and functions. There are two potential candidates for

a square root, and we select one. This is the whole reason Riemann surfaces exist;

to turn expressions that are multifunctions when seen in the complex plane into

functions when defined in Riemann surfaces.

 

[chiro]

No you have it confused: the square root of a complex number has all the properties of a function and thus it only produces one unique value.

The roots of a variable is not the same as taking some nth root of a number: it is completely different.

The function is mapping many things to one thing and that's why when you find the nth roots of a complex number, you get multiple answers because the function that produced them mapped all the inputs to one output.

De-moivre's theorem with regard to finding roots is completely different than using a function mapping to take something from one set and map it to another.

Don't confuse the two.

 

[Bacle2]

I do not believe, nor have I claimed that taking the nth root of a number is equivalent

to the n-th root function.

There is also the fact , re a function:

If a function f maps n-to-1 ,as is the case with the n-th root, then , its inverse (inverse as a relation) maps 1-to-n

and so this inverse cannot be a function-- unless its domain is restricted.

I think you have misunderstood what I'm saying.

 

[Bacle2]

At any rate, I'll explain my layout better:

1)We first define the square root of a (positive for now) number x , to be any number y with y^2=x. This is a 1-to-2

relation. Similar applies to defining n-th root ( which exists for negative numbers when

n is odd, but let's restrict to positive for now ).

2) In order to turn above relation into a function, we select _exactly_ one of the

n-th roots of the number. We define the n-th root _function_ evaluated at

x, to be the specific number y above that satisfies y^n =n . We could have

defined an n-th root function by selecting _precisely_ one other solution to y^n=x.

3) In order to do the selection in 2) , we must restrict the domain , so we can remove

all (n-1) solutions to y^n=x . Otherwise, the relation--a function too, in this case --

will not have an inverse:

If the relation allows n-to-1 , say , the pairs (x1,y),(x2,y),...,(xn,y) are in the

relation, then the pairs : (y,x1), (y,x2),...,(y,xn) are , by definition, part of the

inverse relation --which cannot be a function. z^n is n-to-1 , meaning that its

inverse --as a relation-- is 1-to-n , since if the pairs {(z1,f(z)),(z2,f(z),...(zn,f(z))}

={(z1,y),..,(zn,y)} are in the relation created by z^n , so that the collection of pairs:

{ (y,z1),(y,z2),..,(y,zn)} are in the inverse relation. This inverse relation has to be

restricted to turn it into a function, because we do not allow two different pairs (y,zi),

(y,zj) in a function.

4) Riemann surfaces are used to help avoid n-to-1 relations. This is why many n-to-1 functions

are defined in Riemann surfaces , instead of in the complex plane. There are even examples of

oo-to-1 functions, like logz (note I'm not using caps here , but lower-case instead.) , which are represented

by a "parking lot" Riemann sphere. We then _select_ a branch of logz ; usually the main branch Logz

to turn a 1-to-oo relation into a function.

If you disagree with this , then please explain to me why we define branches of functions in the complex plane.

 

[jgens]

Just my 2 cents here:
1. In the real case when you see a square-root sign it almost always means the principal square-root.
2. In the complex case things get more complicated. The issue now is that while we can define a square-root function on the whole complex plane, there are several ways we can naturally consider doing this and none of these ways make the square-root function continuous. For example
r\exp(i\psi) \mapsto \sqrt{r}\exp(\frac{i\psi}{2}) \;\;\;:\;\;\; -\pi &lt; \psi \leq \pi
r\exp(i\theta) \mapsto \sqrt{r}\exp(\frac{i\theta}{2}) \;\;\;:\;\;\; 0 &lt; \theta \leq 2\pi
Both of these are defined by the same formula, but they define a different function depending on which values of \psi and \theta we allow. It is also worth noting that the first function is discontinuous along the negative real axis while the second function is discontinuous along the positive real axis. To get around this, these functions are often defined without the discontinuous parts. This is one reason why choosing branches in the complex case becomes so important, especially since while there is a principal branch for the square-root in the complex numbers, it is fairly common to use non-principal branches in the complex case (unlike the real case).

 

[jackmell]

I think mathematics would benefit tremendously if students were taught early on that square root and other inverse functions are indeed multivalued. That is I do not feel we do justice to math by viewing \sqrt{z} as a single-valued entity. Granted, when mapped over it's normal Riemann surface, it is single-valued, but it still has two values for each value of z except at the branch-point. So what happens now is that we don't and when all of a sudden the student finds himself in Complex Variable or Analysis, he finds the entire concept of multi-valued functions, branch-cuts, branches, Riemann surfaces, and worst of all, contour integrals over them, a formidable undertaking, gets discouraged, and Mathematics looses another potential genius.

 

[jgens]

I think it depends. While it may make the complex variable theory a bit harder (and I am not convinced that it does), it certainly simplifies the real variable theory to consider only principal branches.

 

[chiro]

We are talking about the set of complex numbers aren't we?

I have a question: you guys are talking about sending re^itheta to r^1/2e^i/2*theta

Now cos(1/2theta) and sin(1/2theta) are both continuous and differentiable with respect to theta. SQRT(r) on the other hand is also since r > 0.

All of these representations: SQRT(r), cos(theta/2) and sin(theta/2) all produce single answers. How is this not a function?

 

[jgens]

How is this not a function?

It is a function. But do you mind telling me which of the functions below you are talking about?
r\exp(i\psi) \mapsto \sqrt{r}\exp(\frac{i\psi}{2}) \;\;\;:\;\;\; -\pi &lt; \psi \leq \pi
r\exp(i\theta) \mapsto \sqrt{r}\exp(\frac{i\theta}{2}) \;\;\;:\;\;\; 0 &lt; \theta \leq 2\pi

 

[chiro]

The bottom one.

 

[jgens]

The bottom one.

The bottom one is actually the non-principal branch. The point I am hoping to get across is that, unlike in the real variable case, there is not a nice canonical choice regarding which branch to work with in the complex case. This was Bacle2s contention about the expression z^1/2 being ambiguous without specifying the branch.

Edit: To be clear, I actually disagree that the expression is ambiguous since I would take z^1/2 to mean any square-root of z in the given context, but I do agree that the expression would be ambiguous if the OP were talking about functions instead of identities for the reasons expressed above.

 

[rbj]

The square root of a non-negative real number is always non-negative.

that is simply not correct.

doesn't matter if it's real or complex, if it ain't zero, there are always two square roots to every number, and they are negatives of each other.

 

[chiro]

The bottom one is actually the non-principal branch. The point I am hoping to get across is that, unlike in the real variable case, there is not a nice canonical choice regarding which branch to work with in the complex case. This was Bacle2s contention about the expression z^1/2 being ambiguous without specifying the branch.

Edit: To be clear, I actually disagree that the expression is ambiguous since I would take z^1/2 to mean any square-root of z in the given context, but I do agree that the expression would be ambiguous if the OP were talking about functions instead of identities for the reasons expressed above.

I'm talking about obtaining just a complex number in the cartesian form by taking the square root of it: you can convert to polar form or whatever, but as long as it's non-zero it exists and is unique.

I am aware of the different branches, but the evaluation of the final quantity will always be unique: all valid functions are.

 

[pwsnafu]

that is simply not correct.

doesn't matter if it's real or complex, if it ain't zero, there are always two square roots to every number, and they are negatives of each other.

There are indeed two square roots, but I used the definite article for a reason. Only the positive one is called "the square root". The square root function on positives is indeed a function.

 

[jgens]

I'm talking about obtaining just a complex number in the cartesian form by taking the square root of it: you can convert to polar form or whatever, but as long as it's non-zero it exists and is unique.

I am aware of the different branches, but the evaluation of the final quantity will always be unique: all valid functions are.

But if you are talking about a square-root function in the complex plane you need to specify which branch, otherwise it is ambiguous which square-root function you are talking about; that is, unlike in the real variable case, it is not sufficient to say let \sqrt{\cdot}:\mathbb{C} \rightarrow \mathbb{C} be the square-root function without specifying the branch.

Now, as far as the OP goes, none of this really matters since the question can be correctly interpreted without the use of functions. But if you are going to start talking about a square-root function in the complex plane, it is best to specify a branch (or work on a Riemann surface to eliminate the multi-valued problem).

 

[Bacle2]

I think mathematics would benefit tremendously if students were taught early on that square root and other inverse functions are indeed multivalued. That is I do not feel we do justice to math by viewing \sqrt{z} as a single-valued entity. Granted, when mapped over it's normal Riemann surface, it is single-valued, but it still has two values for each value of z except at the branch-point. So what happens now is that we don't and when all of a sudden the student finds himself in Complex Variable or Analysis, he finds the entire concept of multi-valued functions, branch-cuts, branches, Riemann surfaces, and worst of all, contour integrals over them, a formidable undertaking, gets discouraged, and Mathematics looses another potential genius.

You may have a point; but it is tricky, I think: you may also overwhelm some with

too-high of a lvel of complexity. Still, along your comment, I think too, that in undergrad.

one learn mostly, if not only, to work in ℝⁿ , which is a space with very

nice natural structure at just-about all levels: metric space, vector space, convex,

complete, etc. And then, when you move on to topology or general metric space

theory, you suddenly find yourself working with bare-bone spaces with very

little structure.