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

[Mark44]

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.

 

[Mark44]

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.

 

[Svein]

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, <br /> \sqrt[3]{-1}=e^{\frac{5i\pi}{3}}=\frac{1}{2}-i\frac{\sqrt{3}}{2}

 

[PeroK]

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, <br /> \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.

 

[wrobel]

Is reaching the second page of posts on whether or not a case of analytic continuation?

It is. Because the correct explanation of these effects is eventually in terms of analytic functions and their branches. As I pointed out previously. To say $\sqrt {-1}=i$ means to fix a branch of $\sqrt z$

 

[FactChecker]

No, both symbols mean the same thing.

IMO, in complex analysis, $z^{1/2}$ is not necessarily assumed to mean the principal square root.

 

[Erik 05]

You're misusing the rules of exponents here. The rules you're using presume that the base is greater than or equal to zero.
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.

$(-4)^{1/2}=\pm\sqrt{-4}=\pm 2i$

$\sqrt 9 = +(9^{1/2})$, which is 3

I don't believe I'm misusing or confusing anything.

 

[Mark44]

$(-4)^{1/2}=\pm\sqrt{-4}=\pm 2i$

I don't believe I'm misusing or confusing anything.

Although there are two square roots of -4, I believe the convention is to evaluate $(-4)^{1/2}$ to its principal value, which is 2i.

 

[Erik 05]

Although there are two square roots of -4, I believe the convention is to evaluate $(-4)^{1/2}$ to its principal value, which is 2i.

Well, that's the convention I use, as I said, less confusing, but at the end of the day it might well come down to whether you are doing maths or physics, or whatever problem you are solving - often you are only interested in positive values, or real numbers, though not always.

 

[PeroK]

Well, that's the convention I use, as I said, less confusing, but at the end of the day it might well come down to whether you are doing maths or physics.

You mean you could do an experiment? Fire a beam of numbers through a square root machine and see whether they come out as one beam of principal square roots, or (like Stern-Gerlach) two beams, each corresponding to a separate root?

 

[Erik 05]

Not sure, would the numbers be entangled? Maybe some quantum computing algorithm...

Ok, I did a physics degree a few years ago, having a background in maths, and I have to say that some of the maths that physicists do did strike me as, well, heresy, almost.

 

[Grasshopper]

This thread has become far more than I had hoped. I've learned a lot more about complex numbers than I anticipated I would. It's also interesting to see the limits of convention and how that messes with people.

Can we have a similar thread with the obelus, and grouping conventions (like mulitplication by juxtaposition)? 8÷2(2 + 2) is the latest trendy one, I believe.

_{(I kid, I kid)}

 

[Svein]

IMO, in complex analysis, $z^{1/2}$ is not necessarily assumed to mean the principal square root.

Which is the point I was trying to make above, but forgot to stress: The concept of "principal root" is meaningless in complex analysis. A (possibly) better example: What is the square root of i? Skipping the small steps of adding multiples of 2πi, we have i=e^{\frac{i \pi}{2}+2n\pi i} and therefore \sqrt{i}=e^{\frac{i\pi}{4}+n\pi i}=(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\cdot (-1)^{n}.

 

[FactChecker]

Which is the point I was trying to make above, but forgot to stress: The concept of "principal root" is meaningless in complex analysis.

I wouldn't say that. The principle root is not assumed, but it is a fundamental concept that is studied thoroughly in complex analysis.

 

[PeroK]

Which is the point I was trying to make above, but forgot to stress: The concept of "principal root" is meaningless in complex analysis. A (possibly) better example: What is the square root of i? Skipping the small steps of adding multiples of 2πi, we have i=e^{\frac{i \pi}{2}+2n\pi i} and therefore \sqrt{i}=e^{\frac{i\pi}{4}+n\pi i}=(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\cdot (-1)^{n}.

Why do you need the $(-1)^n$ if $\sqrt 2 = p(2)(-1)^n$? Where $p(2)$ is the principal square root of $2$?

Also, if $\sqrt 2 = \pm p(2)$, then you now have a problem with the expression $\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$, as this is now a set of four complex numbers, rather than one or two.

Like it or not, you are forced to use the concept of $\sqrt 2$ being a single number - or invent a new notation, such as $p(2)$ to force it to take a single value.

 

[jedishrfu]

It seems we have exhaustively explored all the various viewpoints of this topic and all beginning to reiterate key areas of thought and with that in mind it’s good to tank everyone for contributing and to close this thread now.

Closing now with an excellent summary from @pbuk :

When we are working with complex numbers there are in general 2 solutions to the equation $w^2 = z$.

In some branches and applications of mathematics we don't need to distinguish between the solutions, and so in these contexts when we write $w = \sqrt z$ we mean that $w$ is any number that satisfies the equation $w^2 = z$, for example $w = \sqrt{-4} \implies w \in \{2i, -2i\}$.

In other branches and applications of mathemetics it is useful to define the 'principal square root' such that when we write $w = \sqrt z$ we mean that $w$ is the principal square root of $z$; for example $w = \sqrt{-4} \implies w = 2i$.

When we represent complex numbers as points in the Argand Plane, we define the principal n'th root of a number $\sqrt[n]{z}$ as the solution $(r, \theta)^n = z$ with the smallest positive value of $\theta$. It is worth noting:

• this is compatiable with the definition that is (almost) always used in $\mathbb R$ of the principal square root of a positive number where $x = -2$ is not a solution to $x = \sqrt 4$.
• computer languages and computer algebra packages often implement a complex-valued sqrt or similar function which returns the principal square root according to this definition.

Thank you,
The Mentors