# Regarding i^2, and why it's -1 and not 1

Gold Member
TL;DR Summary
Some clarification on what the positive 1 solution of the root of the products means, and more
________________
So presumably ##x = \sqrt{-1}## can be derived as a solution to the equation
##x^2 + 1 = 0##, thus,
##x = ± \sqrt{-1}##

(1) We always use the positive root. What is the reason for that? Is it simply a convention? Or are there dire consequences for using the negative root, such as causing the universe to spontaneously collapse in an ill-conceived division by zero operation? ;)

(2) Regarding the argument that ##i^2 = (\sqrt{-1})(\sqrt{-1}) = \sqrt{(-1)(-1)} = \sqrt{1} = 1##, is that simply an uninteresting and useless result, and that's why it's never taught? Or is it not allowed mathematically to do this? It seems almost like transforming from the complex axis to the real axis. What does it mean? Is there anything interesting regarding this result?

Thanks!

Not anonymous

Staff Emeritus
Gold Member
The thing that you cannot do is ##\sqrt{a}\sqrt{b}\neq \sqrt{ab}##. It's only true for positive real numbers, as you've observed here.
(1) We always use the positive root. What is the reason for that? Is it simply a convention? Or are there dire consequences for using the negative root, such as causing the universe to spontaneously collapse in an ill-conceived division by zero operation? ;)

What makes you think we are not using the negative root? ##i## and ##-i## are algebraically indistinguishable - you can map them to each other and nothing really changes.

PeroK and Grasshopper
Gold Member
The thing that you cannot do is ##\sqrt{a}\sqrt{b}\neq \sqrt{ab}##. It's only true for positive real numbers, as you've observed here.

What makes you think we are not using the negative root? ##i## and ##-i## are algebraically indistinguishable - you can map them to each other and nothing really changes.
Wait, how? Would you mind elaborating on this?

(side note: would this possibly be why special relativity can be done with either (+---) or (-+++) sign conventions?)

Staff Emeritus
Gold Member
Wait, how? Would you mind elaborating on this?

Consider the map ##\sigma:\mathbb{C}\to \mathbb{C}## defined by ##\sigma(a+bi)=a-bi##. This is a field automorphism - it's a bijection for which ##\sigma(w+z)=\sigma(w)+\sigma(z)##, and ##\sigma(wz)=\sigma(w)\sigma(z)##. You can think of ##\sigma## as being a function that asks "what changes if we used ##-i## instead of ##i##" and the fact that it preserves addition and multiplication means the answer is "not a lot".

In general, ##i## is just defined to be an arbitrary choice of one of the square roots of -1, and it didn't matter which one you pick (in fact, there isn't even a way to pick one of them vs the other)
(side note: would this possibly be why special relativity can be done with either (+---) or (-+++) sign conventions?)

The main point is special relativity is that the metric is conserved. If ##f(t,x)=t^2-x^2## is constant in all frames, then trivially so is ##g(t,x)=-f(t,x)##.

pbuk and Grasshopper
Gold Member
We always use the positive root. What is the reason for that? Is it simply a convention?
yes.

Delta2
Homework Helper
Gold Member
With only positive numbers ## a\gt 0##, ##\sqrt{a}## is always considered the positive number by convention. With negative or complex numbers, ##z##, the square roots and powers should be handled more carefully using arg(##z##). When that is done, is never true that ##i^2 = 1##.

Grasshopper
Using only positive is a convention. ##\sqrt{-1}=\pm i## is mathematically correct.

Homework Helper
Gold Member
Using only positive is a convention. ##\sqrt{-1}=\pm i## is mathematically correct.
Not this again. This is NOT correct. The symbol ##\sqrt w ## denotes the principal square root function which has a single value (it is a function), which when ## w = -1 ## we call ## i ##.

This has nothing to do with the fact that the selection of ## i ## is entirely arbitrary, and yes if we set ## j = -i = - \sqrt{-1} ## then we can rewrite the whole of complex arithmetic using ## j ## instead of ## i ## and it will still work, but that does not make ## j = i ##, nor does it apply the ## \pm ## alternative to the square root symbol to stop it representing a function.

https://mathworld.wolfram.com/SquareRoot.html

spinam, Delta2, Mark44 and 1 other person
Homework Helper
Gold Member
(in fact, there isn't even a way to pick one of them vs the other)
This.

2022 Award
(side note: would this possibly be why special relativity can be done with either (+---) or (-+++) sign conventions?)
If you're working from a book that's using ##i## in special relativity, I recommend dropping the book at least as far as teaching relativity is concerned (yes, even if it's Feynman's Lectures). It's not a helpful convention. Once you understand relativity without it, then you can go back and read sources that use it if you like.

The sign convention in the line element in relativity, though, doesn't have anything to do with ##i##. It's just that you can always have overall positive, zero, and negative squared "lengths" whatever convention you use, so people use the convention that's more convenient in their field (or the one their professor used...).

Homework Helper
This seems strangely contentious, but I agree with people who say there is no such thing as a "principal" square root of -1. More easily understood perhaps is the fact that there is no positive square root of -1. I.e. there is no ordering on the complex numbers, no division into positive and negative. (If i were positive then i^2 = -1 would also be positive.) And the symmetry of the complex numbers means, as stated above, there is no way to distinguish one square root of -1 from another.

This is confusing, since if you think the complex numbers are naturally represented as pairs of reals, then you may think that the complex number (0,1) is more positive or more principal than (0,-1), but this is an illusion of choosing this representation. For example, suppose I decide to write complex numbers as a+bj, instead of a+bi, and I define my j to equal your -i. Then j looks principal in my system.

The same confusion occurs on a more elementary level when people call -x "negative x", rather than "minus x". I.e. if x is a negative number then -x is positive. I.e. many people think when they see the symbol "-" in front of another number, that they are looking at a negative number. "minus" means additive opposite, "negative" means less than zero. But I don't expect to convince the skeptical about this.

spinam
Gold Member
This seems strangely contentious, but I agree with people who say there is no such thing as a "principal" square root of -1.
Of course there are multiple roots in the complex plane. There is a "principle root" if the person you are talking to has chosen to define one. Which we all usually do for convenience.

Furthermore, you are much more likely to see the branches 0 ≤ arg(z) < 2π, and -π < arg(z) ≤ π, than 2π ≤ arg(z) < 4π. As soon as you hear the phrase "principle root" then you know what's going on and you have about a 50% chance of knowing the branch they are talking about.

So yes, in practice a principle root exists, because it's in common use. That doesn't mean that we all don't understand there are other possibilities. This has more to do with how humans communicate Math than the actual Math.

pbuk
Gold Member
Of course there are multiple roots in the complex plane. There is a "principle root" if the person you are talking to has chosen to define one. Which we all usually do for convenience.

Furthermore, you are much more likely to see the branches 0 ≤ arg(z) < 2π, and -π < arg(z) ≤ π, than 2π ≤ arg(z) < 4π. As soon as you hear the phrase "principle root" then you know what's going on and you have about a 50% chance of knowing the branch they are talking about.

So yes, in practice a principle root exists, because it's in common use. That doesn't mean that we all don't understand there are other possibilities. This has more to do with how humans communicate Math than the actual Math.
Interesting you should say this, because I was going to ask if this situation is much different from how you’ll see something involving ##\frac{nπ}{L} ## when solving a heat equation.

Homework Helper
@DaveE: In my opinion, you make an "error" in assuming the usual "complex plane" is a canonical model of a field algebraic over the reals, in which there is a square root of -1, i.e. of the "complex numbers" The complex plane you are using is more than a copy of the algebraic closure of the reals. It is a copy of that algebraic closure plus a choice of orientation, i.e. a choice of principal square root of -1. Thus, to me, the so called principal root you refer to is an artifact of the concrete model you have chosen for the complex numbers. It is not defined in field theoretic terms.

For example, if C and D are both copies of the complex numbers, and C-->D is a field isomorphism, then it should send the "principal" square root of -1 in C to the principal square root of -1 in D. But this is not always true. In contrast, if R and S are two copies of the real number field, then any field automorphism from R to S will always send the principal square root of 2 in R to the principal square root of 2 in S.

Perhaps a happier resolution of this artificial linguistic controversy would be for me to accept your use of this word, but just to remind you that isomorphisms of the complex field do not preserve your "principal" square roots. This makes them rather less than useful in field theoretic considerations, at least in my opinion. I.e. I might concede there is a principal square root of -1 in the usual complex plane in some sense, but it has no field theoretic significance.

As usual, these disagreements hinge on different understandings of the meanings of words. You are assuming the complex numbers are defined as pairs of reals, and in that setting, you can distinguish a particular square root of -1, namely as (0,1). I am assuming the right definition is as the algebraic closure of the reals, and there is no way to single out a particular square root in that setting.

I.e. assume F is an algebraic closure of the reals and that is all. I.e. F is an algebraically closed field and is algebraic over the reals. Then how do you define the argument of an element of F? All you know is that F is 2 dimensional as a vector space over R, but you are not given a particular basis.

So I think we don't really disagree on the substance, just on what properties of the complex numbers we consider as intrinsic, when defining the term "principal square root". So I guess it is unimportant. But I did enjoy thinking abnout it. Thank you!

DaveE
Staff Emeritus
Gold Member
Maybe the real takeaway is that canonical representations can be overrated sometimes, and a good non canonical representation of an object can have some highly powerful value.

DaveE
Homework Helper
Yes indeed. I myself am a complex analytic geometer, and for that subject, holomorphic, i.e. orientation preserving, maps of the complex plane are crucial.

So this confusion seems to stem from different perspectives. If we think of the complex numbers C as ordered pairs of reals, then it is possible to distinguish a preferred square root of -1, namely (0,1), and equivalently there is given a preferred (ordered) basis of C over R, namely (1,0) and (0,1). But if we just assume C is an algebraically closed field extension algebraic over R, there isn't any such additional data given.

If points up again for me, the importance of giving definitions. Namely what is a "principal" square root, i.e. what properties should it have? Once these are specified, one can check them in given cases.

I apologize for assuming other people were in error, when they were only using different assumptions.

Homework Helper
Gold Member
Principle branches of the logarithm, roots, and many other functions are very fundamental in complex analysis and have been standard for a hundred years.

Gold Member
Maybe the real takeaway is that canonical representations can be overrated sometimes, and a good non canonical representation of an object can have some highly powerful value.
Exactly. And when that's the case the author needs to recognize that some readers will use the common assumptions and be confused. If you need to deviate from the common shortcuts that are out there, you would do well to tell your readers before they misunderstand.

FactChecker
Homework Helper
@FactChecker: Of course this is true. Thank you for provoking me to puzzle out why this is so, when in the purely algebraic sense, there is no such principal root. You have made see more clearly that the familiar complex plane, so useful in analysis, is slightly more than a choice of algebraic closure of the reals, but is such a closure together with a preferred square root of -1, which is equivalent geometrically to a choice of orientation of that 2 dimensional real vector space. I never realized the truth of this principle before, i.e. that the usual complex plane includes some extra data, not given by the notion of algebraic closure. In fact I am more familiar with the analytic version, and have often been puzzled in doing abstract field theory, when the author makes a point to remind me that in that setting no such preferred roots exist.

FactChecker
Homework Helper
Gold Member
@FactChecker: but is such a closure together with a preferred square root of -1, which is equivalent geometrically to a choice of orientation of that 2 dimensional real vector space.
I think you are correct. The geometric applications require a distinction between a 90-degree counterclockwise rotation from the positive reals versus a clockwise rotation. There may be other reasons, but this is one that I am aware of.

cmb
Is the question? ...If SQRT(-1) = ±i , then can one multiply both roots together to get (-i)*(+i) = -(-1) = 1 ?

That'd be like saying SQRT(9) = ±3, so multiply those roots to get 9 = (-3) * (+3) = -9

... I guess 'conventions' are needed sometimes ...

I was always taught to figure out both roots then make sure the answer made sense. Curiously, I have never stopped to question that 'obvious' piece of wisdom. But obviously there are limits to working with all roots of any polynomial. Are there any 'rules' that emerge from the mathematics, or do we have to stick with '...just check if it makes sense ...'?

Homework Helper
Gold Member
Are there any 'rules' that emerge from the mathematics, or do we have to stick with '...just check if it makes sense ...'?
If you carefully keep track of all the restrictions, then you can probably get only the right answers. But often it is much easier and more practical to just get a set of possible answers and check which really work. Even if you try to keep track of all the restrictions, it is wise to check the answers. Some might not work, in spite of your best efforts. Also, you might have made a mistake that eliminated some of the valid answers.

Erik 05
I would use (-1)1/2, since I'm not exactly sure what sqrt(-1) means.

Mentor
I would use (-1)1/2, since I'm not exactly sure what sqrt(-1) means.
##\sqrt {-1}## means the number, when squared, that results in -1.

pbuk
Summary:: Some clarification on what the positive 1 solution of the root of the products means, and more
________________

We always use the positive root.
Summary:: Some clarification on what the positive 1 solution of the root of the products means, and more
________________

Regarding the argument that , is that
it is not an argument

Actually these things are already addressed to nontrivial theory of analytic functions, branches of analytic functions etc it is essentially beyond basic complex numbers theory.

You just should know that ##\sqrt{-1}=\{\pm i\}## and the expression ##\sqrt{-1}\cdot \sqrt{-1}## is not defined correctly.
From the viewpoint of analytic functions theory the expression ##\sqrt 1\cdot \sqrt 1## is also not correctly defined since ##\sqrt{1}=\{\pm 1\}##

PeroK
Erik 05
##\sqrt {-1}## means the number, when squared, that results in -1.
You sure about that, the number? or numbers? It's used both ways, although I'm pretty sure it's defined as principle root, which is why you usually see a +/- before it.

So that makes you half wrong. (-1)^1/2 is less ambiguous.

Homework Helper
Gold Member
2022 Award
Is reaching the second page of posts on whether ##\sqrt {-1} = i## or not a case of analytic continuation?

pbuk, vela and Grasshopper
Gold Member
I would use (-1)1/2, since I'm not exactly sure what sqrt(-1) means.
Aren’t those two symbols for the same thing?

Erik 05
sqrt(-1) is one root (a function), (-1)^1/2 = +/- sqrt(-1) is both roots (not a function) ...as far as I know, I could be wrong. Computers treat them the same, but they are dumb. My thinking is it's direct from x2 = -1 --> x(2 x 1/2) = (-1)(1 x 1/2) ---> x = (-1)1/2. Not specifying a specific root there.

Last edited:
cmb
sqrt(-1) is one root (a function), (-1)^1/2 = +/- sqrt(-1) is both roots (not a function) ...as far as I know, I could be wrong. Computers treat them the same, but they are dumb. My thinking is it's direct from x2 = -1 --> x(2 x 1/2) = (-1)(1 x 1/2) ---> x = (-1)1/2. Not specifying a specific root there.
It's not a convention I have seen before, never thought they were different. But conventions are things that are adopted, habitually, over time, so stuff like that can change?

Mentor
You sure about that, the number? or numbers? It's used both ways, although I'm pretty sure it's defined as principle root, which is why you usually see a +/- before it.
You're saying two conflicting things -- principal root means one root, not two.
This is why we say, for example, that ##\sqrt 4 = 2## rather than ##\sqrt 4 = \pm 2##
So that makes you half wrong. (-1)^1/2 is less ambiguous.
No, both symbols mean the same thing.

Mentor
sqrt(-1) is one root (a function), (-1)^1/2 = +/- sqrt(-1) is both roots (not a function) ...as far as I know, I could be wrong.
You're misusing the rules of exponents here. The rules you're using presume that the base is greater than or equal to zero.
Computers treat them the same, but they are dumb. My thinking is it's direct from x2 = -1 --> x(2 x 1/2) = (-1)(1 x 1/2) ---> x = (-1)1/2. Not specifying a specific root there.
You are confusing two concepts here: solving a quadratic equation (which generally has two roots), and taking the square root of both sides of an equation.

This is the difference between solving the equation ##x^2 = 9##, with roots x = 3 or x = -3 versus evaluating ##\sqrt 9 = 9^{1/2}##, which is 3.

Written another way: $-1=1*e^{i\pi}=e^{i\pi}=e^{3i\pi}=e^{5i\pi}\dotso$. So, taking the square root: $\sqrt{-1}=\sqrt{1\cdot e^{i\pi}}=\sqrt{1}\cdot \sqrt{e^{i\pi}}=1\cdot e^{i\frac{\pi}{2}}=e^{i\frac{\pi}{2}} =i$. But also $\sqrt{-1}=e^{\frac{3\pi i}{2}}=-i$ and so on.

As an aside: This method makes finding the third and fourth roots easy. For example: $\sqrt[3]{-1}=\sqrt[3]{e^{i\pi}}=\sqrt[3]{e^{3i\pi}}=\sqrt[3]{e^{5i\pi}}$. From this basic setup you get $$\sqrt[3]{-1}=e^{\frac{i\pi}{3}}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$$, $$\sqrt[3]{-1}=e^{\frac{3i\pi}{3}}=-1$$, $$\sqrt[3]{-1}=e^{\frac{5i\pi}{3}}=\frac{1}{2}-i\frac{\sqrt{3}}{2}$$

spinam and Grasshopper
Homework Helper
Gold Member
2022 Award
Written another way: $-1=1*e^{i\pi}=e^{i\pi}=e^{3i\pi}=e^{5i\pi}\dotso$. So, taking the square root: $\sqrt{-1}=\sqrt{1\cdot e^{i\pi}}=\sqrt{1}\cdot \sqrt{e^{i\pi}}=1\cdot e^{i\frac{\pi}{2}}=e^{i\frac{\pi}{2}} =i$. But also $\sqrt{-1}=e^{\frac{3\pi i}{2}}=-i$ and so on.

As an aside: This method makes finding the third and fourth roots easy. For example: $\sqrt[3]{-1}=\sqrt[3]{e^{i\pi}}=\sqrt[3]{e^{3i\pi}}=\sqrt[3]{e^{5i\pi}}$. From this basic setup you get $$\sqrt[3]{-1}=e^{\frac{i\pi}{3}}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$$, $$\sqrt[3]{-1}=e^{\frac{3i\pi}{3}}=-1$$, $$\sqrt[3]{-1}=e^{\frac{5i\pi}{3}}=\frac{1}{2}-i\frac{\sqrt{3}}{2}$$
We're just going round in circles with some people believing that ##\sqrt z## is uniquely defined as a principal square root, and others saying that ##\sqrt z## is a set of two numbers.

For those that believe the latter, however, you shouldn't need to write ##\pm\sqrt {z}## as in: $$z = \frac{-b \pm \sqrt {b^2 - 4ac}}{2a}$$ But, instead have: $$z = \frac{-b + \sqrt {b^2 - 4ac}}{2a}$$ if ##\sqrt z## is indeed multi-valued.