Complex or not?

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jeffreyH

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Complex or not?
« on: 06/01/2015 23:55:21 »
If we have the relationship x^2 = -25 then x = SQRT(-25) which is equivalent to x = 5i giving (5i)^2 = -25. Could another form of the complex relationship be described as -1*(x^2) = -25 so that like i being separated out we can use -1 as a post multiplier? Then -1*x = -5 becomes another representation so that -1 becomes implicit in the definition of imaginary numbers in a slightly different way. Would this make any difference to calculations and the solutions of equations?

PmbPhy

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Re: Complex or not?
« Reply #1 on: 07/01/2015 05:20:33 »
If we have the relationship x^2 = -25 then x = SQRT(-25) which is equivalent to x = 5i giving (5i)^2 = -25. Could another form of the complex relationship be described as -1*(x^2) = -25 so that like i being separated out we can use -1 as a post multiplier? Then -1*x = -5 becomes another representation so that -1 becomes implicit in the definition of imaginary numbers in a slightly different way. Would this make any difference to calculations and the solutions of equations?
Actually if you have x^2 = - 25 it follows that the complete solution is

$$x = 5ie^{\theta}$$

Where the exponent in the e is arbitrary. This is crucial to know in quantum mechanics since when you add two solutions these phase factors must be chosen by initial conditions.
« Last Edit: 07/01/2015 17:49:17 by PmbPhy »

evan_au

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Re: Complex or not?
« Reply #2 on: 07/01/2015 08:52:47 »
Quote from: jeffreyH
-1*(x^2) = -25

This is the same as solving x2 = -25* x2 = 25, for which the solution is the simple x=5, or the slightly less obvious x=-5.

Perhaps you meant y2*x2=-25, where y2=-1?
In this case, y=i (or -i), and x=5 (or -5).

I recall from maths class that when you solve an equation like xn=-1, the solutions are positioned on a circle of radius 1 on the complex plane, with the n solutions equally spaced around the unit circle.

To add to the comment from PmbPhy, this comes about because of Euler's formula:
x=e = cos(θ) + i sin(θ)

and

xn=einθ = cos(nθ) + i sin(nθ) =-1

If you want x2=-25, then the circle has radius 5.

When n=2, there are two solutions, positioned opposite each other on this circle.

Quote
another representation so that -1 becomes implicit in the definition of imaginary numbers in a slightly different way
There is a generalisation of imaginary numbers, called quaternions.
Quaternions extend complex numbers, similar to the way that complex numbers extend real numbers.

In this representation, there are three different square roots of -1: i, j & k.
i2 = j2 = k2 = ijk = −1
where: i ≠ j ≠ k ≠ i

*Correction from PmbPhy - thanks!
« Last Edit: 07/01/2015 20:05:34 by evan_au »

jeffreyH

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Re: Complex or not?
« Reply #3 on: 07/01/2015 12:26:44 »
Thanks for the answers. Evan I thought it was the case that quarternions were out of favour and had been for a long time.

PmbPhy

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Re: Complex or not?
« Reply #4 on: 07/01/2015 15:11:52 »
Quote from: evan_au
Quote from: jeffreyH
-1*(x^2) = -25

This is the same as solving x2 = -25, for which the solution is the simple x=5, or the slightly less obvious x=-5.
Why do you say that? The answer is complex whereas you have x = 5 which is real. That is to say, you're confusing this with x^2 = 25.

Jeff got that from solving

x^2 = -25

by taking the square root of both sides to get

sqrt[x^2] = sqrt[-25] = sqrt[-1]sqrt[25]

Since sqrt[x^2] = |x|, i = sqrt[-1] and sqrt[25] = 5 we get

|x| = 5i

We get rid of the absolute sign by introducing a phase to get what I got.

lightarrow

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Re: Complex or not?
« Reply #5 on: 07/01/2015 15:48:52 »
x2 + 25 = 0 has the solutions:

x = 5e, θ = π/2 + k*π; k is an integer number.

for k = 0: x =5i, because eiπ/2 = i (Euler's formula).

for k = 1: x = -5i, because e3iπ/2 = -i.

the other values of k gives, periodically, the same two up here.

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« Last Edit: 07/01/2015 15:56:13 by lightarrow »

lightarrow

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Re: Complex or not?
« Reply #6 on: 07/01/2015 15:54:59 »
Actually if you have x^2 = 25 it follows that the complete solution is
$$x = 5ie^{\theta}$$
It's a typo, you certainly intended $$x = 5e^{i\theta}$$.

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jeffreyH

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Re: Complex or not?
« Reply #7 on: 07/01/2015 16:36:34 »
x2 + 25 = 0 has the solutions:

x = 5e, θ = π/2 + k*π; k is an integer number.

for k = 0: x =5i, because eiπ/2 = i (Euler's formula).

for k = 1: x = -5i, because e3iπ/2 = -i.

the other values of k gives, periodically, the same two up here.

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lightarrow

I think Richard Feynman summed it up by calling Euler's formula the jewel of physics.

evan_au

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Re: Complex or not?
« Reply #8 on: 07/01/2015 20:00:02 »
Quote from: PmbPhy
Why do you say that?
Oops!  I meant x2 = 25 (Copy/Paste error).
Sorry about that... (correction marked up above).

evan_au

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Re: Complex or not?
« Reply #9 on: 07/01/2015 20:44:58 »
Quote
I thought it was the case that quarternions were out of favour
That is true - each time you generalise numbers beyond imaginary numbers, you lose some useful mathematical properties.

There are certain familiar mathematical properties we had drilled into us at primary school like:
• a+0=a
• a*1=a
• a+b=b+a
• a*b=b*a (the "law" of commutivity)
• c*(a+b)=c*a+c*b
• They apply for real (1 dimensional) & imaginary (2-dimensional) numbers
• ...but they progressively lose validity as you move into quaternions (4-dimensional) and octonions (8 dimensional)
• There are many number systems used by mathematicians, but the thing I find freaky is that these are the only 4 mathematical spaces which behave in this way (you can't do similar mathematics in 3 dimensions, 5 dimensions or 15 dimensions!)
For example, multiplying by two complex numbers can be seen as doing a rotation in a 2-dimensional space (the complex plane). It doesn't particularly matter whether you rotate by θ1 or θ2 first.

However, multiplying by two quaternions can be seen as a rotation in 4-dimensional space. Here it matters greatly whether you rotate first by θ around the i axis or by φ around the j axis! So in this case, a*b ≠ b*a (ie commutivity fails in a 4-dimensional space).

Sometimes problems in the real world really do need only 3, 5, 15 or 10000 dimensions, and these 4 spaces with their special mathematical properties don't apply. So mathematicians and physicists use a more general technique of matrices. The properties of matrices are even less mathematically "nice" than quaternions. For example, in matrices, a*b may not even be defined, and even if it is, usually a*b ≠ b*a. But despite their shortcomings, matrices come in all shapes and sizes, and computers can do operations on them quite efficiently to produce computer games, weather simulations, etc.

But these higher dimensional numbers like quaternions are not totally dead; matrices don't have to contain real numbers - for example, matrices of imaginary numbers can be used for circuit simulation. I have even seen octonions used in theories about the fundamental nuclear particles. It has been suggested that some of the characteristics of 4 dimensional space or string theory are based on the characteristics of these higher-dimensional numbers.[/list]
« Last Edit: 07/01/2015 21:55:45 by evan_au »

jeffreyH

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Re: Complex or not?
« Reply #10 on: 08/01/2015 00:10:22 »
The quaternion group is also cyclic in its power series which is interesting. I knew i was but the fact that the group is sounds very useful. I had not heard of the 4 dimensional space but had been investigating the consequences of complex numbers on geometry. The one thing I don't get is how i, j and k can all be square roots of minus 1 and yet be different. What am I missing? Are there any online explanations?
« Last Edit: 08/01/2015 00:12:17 by jeffreyH »

jeffreyH

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Re: Complex or not?
« Reply #11 on: 08/01/2015 00:32:10 »
I have just looked at the cayley table for the Q group at http://en.wikipedia.org/wiki/Quaternion_group and it doesn't look like associativity holds for this group. My understanding was that was one of the axioms of a group.

Correction: I made an error in the working out. It does look associative.
« Last Edit: 08/01/2015 00:42:12 by jeffreyH »

lightarrow

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Re: Complex or not?
« Reply #12 on: 08/01/2015 22:10:54 »
x2 + 25 = 0 has the solutions:

x = 5e, θ = π/2 + k*π; k is an integer number.

for k = 0: x =5i, because eiπ/2 = i (Euler's formula).

for k = 1: x = -5i, because e3iπ/2 = -i.

the other values of k gives, periodically, the same two up here.

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lightarrow

I think Richard Feynman summed it up by calling Euler's formula the jewel of physics.
It's even more impressive if written:

e + 1 = 0

as Feynman itself writes (if I remember correctly)

which puts in relationship all the 5 "most important" numbers of mathematics.
I have always wondered if there is a similar equation involving a sixth further "important" number  []

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jeffreyH

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Re: Complex or not?
« Reply #13 on: 08/01/2015 23:15:25 »
$$\frac{1}{e^{i\sqcap} + 1}$$

I hate latex.

But this is cool.

http://en.wikibooks.org/wiki/LaTeX/Mathematics#Symbols
« Last Edit: 08/01/2015 23:33:55 by jeffreyH »

jeffreyH

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Re: Complex or not?
« Reply #14 on: 11/01/2015 02:58:20 »