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

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In summary, the use of ##\sqrt{-1}## as a solution to the equation ##x^2 + 1 = 0## is arbitrary and either the positive or negative root can be used. The convention is to use the positive root, but this does not have any mathematical significance. In complex numbers, the symbol ##\sqrt w## represents the principal square root function, which has a single value and is not equivalent to using the ##\pm## notation. The use of ##i## in special relativity is not a helpful convention and can be dropped without affecting the validity of the theory. The sign convention in the line element in relativity is simply a matter of convenience and personal
  • #36
Mark44 said:
No, both symbols mean the same thing.
IMO, in complex analysis, ##z^{1/2}## is not necessarily assumed to mean the principal square root.
 
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  • #37
Mark44 said:
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.
 
  • #38
Erik 05 said:
##(-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.
 
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  • #39
Mark44 said:
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.
 
  • #40
Erik 05 said:
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?
 
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  • #41
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.
 
  • #42
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)? :woot: 8÷2(2 + 2) is the latest trendy one, I believe.

(I kid, I kid)
 
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  • #43
FactChecker said:
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 [itex]i=e^{\frac{i \pi}{2}+2n\pi i} [/itex] and therefore [itex]\sqrt{i}=e^{\frac{i\pi}{4}+n\pi i}=(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\cdot (-1)^{n} [/itex].
 
  • #44
Svein said:
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.
 
  • #45
Svein said:
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 [itex]i=e^{\frac{i \pi}{2}+2n\pi i} [/itex] and therefore [itex]\sqrt{i}=e^{\frac{i\pi}{4}+n\pi i}=(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\cdot (-1)^{n} [/itex].
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.
 
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  • #46
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
 
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