| John
Tiffany |
Hello, I am 55 years old and I already know the square roots of imaginary one (i). But I am going crazy trying to figure out what are the square roots of negative imaginary one. Are they numbers on the complex plane? Or do we have to imagine a third number dimension, j? |
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| Andre
Rzym |
They are indeed numbers on the complex plane. We do not have to imagine a third 'dimension'. Are you comfortable evaluating the square of ? If so, what do you think it is? There is a bit more to say on this topic, but let's start here! Andre |
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| Hauke
Worpel |
John, Let a= i1/2 . Assume a is in the complex plane, so it is of the form a = x+iy. (x+iy)2 =i x2 +2ixy-y2 =i So far so good? x2 =y2 , 2xy=1 (separated out the real and imaginary parts) ±x=±y, xy=1/2 x=y (because from xy=1/2, the sign of y must be the same as the sign of x) x2 =y2 =1/2 x=y=1/sqrt(2) Therefore sqrt(i)=1/sqrt(2)+i/sqrt(2) or the negative of that. Square it out to check. There's easier ways to do this, believe me, but it requires some concepts you haven't met yet. |
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| Andre
Rzym |
Hauke, Unless I am misreading the question, John is asking for 'square roots of negative imaginary one'. In other words . Andre |
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| Stephen
Burgess |
John, There is a result which is proved somewhere in the second year of most university courses called the "Fundamental Theorem of Algebra" (FTA). It states (in slightly more rigorous language), that it you have a polynomial with complex coefficients, then all the roots of it are within the complex plane, and there are no others, so any polynomial of degree n will have n complex solutions (possibly with repeats). The consequences of this are very deep. In your example, you are searching for solutions to the polynomial: z²=-i In other words, what number squared gives the result -i. What the FTA tells us is that this will be in the complex plane. In addition, any other number you try to construct, like the 5th root of 5-2i, will also be in the complex plane. So we will never need a 3rd dimension so long as we start with only 2. |
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| Stephen
Burgess |
(Second half of the post) However, two dimensions does not have to be the limit, so long as we start off in more than two dimensions. It is impossible to construct a self-consistent algebraic structure in 3D, but one can in 4D, provided we don't mind breaking the law of commutativity. The result is the quaternion group, which has applications in theoretical physics. The rules are as follows: i^2=j^2=k^2=-1, ij=-ji=k, jk=-kj=i ki=-ik=j If you wants, you can construct algebraic systems in higher dimensions, but it is necessary to break even more rules of algebra to do this! |
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| Hauke
Worpel |
Sorry Andre, I misread the question. Stephen- I take it that odd-numbered dimensions do not have consistent algebraic structures but even-numbered ones do? Or is it to do with powers of 2? Also, don't those Quaternion rules imply that ijk=-1? |
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| Stephen
Burgess |
Yes, those rules do imply that ijk=-1. It's about powers of 2 rather than even numbers. Octonions also exist, but beyond there I don't think any other structure is useful. Here is a nice place to start if you are still interested, as you are reaching the limit of what I know about these things! |
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| John
Tiffany |
Thanks for all the information, fellows. But I still do not know what is the square root of negative imaginary one, although it is reassuring to know it is in the complex plane. Stephen, you say that quaternions and octonions are useful. Could you mention a possible use? Hauke, how did you get: from x2+2ixy-y2=i (which I can understand) to x2=y2, 2xy=1 ? |
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| John
Tiffany |
Oops, I retract that question to Stephen. Sorry, I had not looked at the article by Girish Joshi when I wrote that. |
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| Andre
Rzym |
It is what I tried to hint that it was above, i.e. Andre |
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| Stephen
Burgess |
> how did you get: from x2+2ixy-y2=i (which I can understand) to x2=y2, 2xy=1 ? < If you take the real part of both sides of the equation, then you get: Re(LHS)=x²-y² Re(RHS)=0 So x²-y²=0 and x²=y². Similarly with the second equation, expect that here we take imaginary parts. |