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why don't imaginary numbers make so much sense?

Discussion in 'Mathematics and Physics' started by PG1995, Aug 29, 2011.

  1. PG1995

    PG1995 Active Member

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    Hi :)

    One of founding pillars of mathematics is that you cannot divide by zero; it's utterly absurd. Therefore, if one starts claiming that division by zero isn't absurd because it pops up while solving some problems. I think then it would cast a lot of doubt on the credibility of mathematics and give rise to disbelief in mathematics.

    Another one of the founding principles of mathematics is that when you multiply any two 'real' numbers you will get a +ve number;no matter even if the numbers were -ve. Now when we write sqrt(-1), we are trying to do something which isn't allowed. I think I don't have much problem with writing swrt(-1) and calling it iota. Obviously, I would have serious problem accepting the result if it was said that sqrt(-1) equals some real number. So, I think my problem only lies in the fact that how come we end up with sqrt(-1) expression while solving other 'normal' problems. How does nature make use of such 'nonsense' expression? Could you please give me some simple example where iota is used and we can make some sense out of it? I don't think nature can make much use iota when one can't even tell which one the two or more imaginary numbers is greater; e.g. you can't tell whether 4i is greater than 2i or not! Please don't use more math to explain math. Thank you.

    Regards
    PG
     
  2. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Well, It's extremely simple to explain it. It adds a "Y" dimension to the classic number line. i or if your in the EE world, you use j as the sqrt(-1). Positive coefficients to j is UP and negative coefficients of J is down.

    Whether 4i is > 2i is easy. What happens is you get numbers such as 2 +3i. That's nothing but a vector. 2 on the x axis and 3 on the Y axis. It's magnitude is computed by c^2 = a^2 + b^2 or C= sqrt (4^2 -9) or C = 16-9 or 7

    The magnitude of the number is just the length.

    It turns out that the impedance of Inductance and capacitance can be written in complex form: Zc=1/jwC and Zl = -JwL

    w is Omega and is the radian frequency where w=2*PI*f where f is the frequency.

    So, now you can visualize an RCL circuit at a given frequency as a varying vector.

    You can convert that number to a polar representation as well if you know the frequency. It can convert to a magnitude @ at a phase angle.

    See how easy that was?

    -j * -j is really the dot product of the two vectors [ 0, -j] and [0, -j] which is 0*0 + -1*-1 or 1.
     
    Last edited: Aug 30, 2011
  3. Ratchit

    Ratchit Well-Known Member

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    PG1995,

    Whoever told you that? Division by zero results in ± ∞, depending on the sign of the dividend. Perhaps you are confusing 0/0 which is undefined.

    Whether it shows up in problems or not is irrelevant. Division by zero has an infinite value.

    It has nothing to do with the credibility or belief of mathematics.

    What is "±ve"? Anyway, it is easily proven that the result of multiplication stays within the same defined number class, and therefore it is not a "founding" principle.

    Whoever told you that? Since when has finding the square root of a number disallowed? I think you are confused. First of all, many math books say or imply that a root of a negative number has a complex esoteric value not conceivable in the physical world. Whereas in fact, complex numbers have a finite value. They should be called "duplex" numbers instead of "imaginary" numbers. Duplex numbers have a real part and an orthogonal part. The symbol "i" or "j" does not mean √-1. "i" or "j" is a mathematical operator, not a finite value. For instance, 7i does not mean i + i + i + i + i + i + i. It means instead, "perform the mathematical operation of rotating the number 7 by 90 counterclockwise (CCW). It is true we get correct results by treating i or j as an arithmetic constant, but that only works because of its conformal similarity.

    So what is the finite value of √-1? In polar form, -1 = 1/_180. Taking the square root we get √-1 = √1/_180 = 1/_90 or 1 rotated by 90 degrees CCW. This number has a real part of 0 and an orthogonal part of +1. It is a duplex number because it takes two values to describe it. Oftentimes, i or j is used as a shorthand for the finite value of 1/_90 or √-1, but it is really a mathematical operator.

    Ratch
     
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  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    What do you mean you cant tell if 4i is greater than 2i or not? Why not?

    Complex numbers (complex numbers have a 'real' part and an 'imaginary' part) can be used to solve more complex problems in ways that work out simpler than other ways. The math behind them is not the same as more ordinary real life math, that's all. It's a different kind of math. You'll run into other types of math sometimes called other 'algebras' when you start to try to solve more types of problems that come up in life especially in engineering.
    Another type of math comes up with working with matrixes. The orientation of two numbers during multiplication is not usually important, such as a*b=b*a, but in matrix math it's more important so a*b might not be equal to b*a.
     
  6. panic mode

    panic mode Well-Known Member

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    common place to use imaginary numbers is when solving problems. one example of such use is solving AC circuits etc.

    real numners can be used to describe only one dimension, such as length for example.
    if two sticks are 1.2 and 1.4m long, it is easy to say which one is longer. note that here thichness of sticks was neglected.

    but for many problems, we cannot just neglect one or two dimensions. this is why we use other representations. most general is vectors. for example vector in 3D would be represented as:
    r=5i+3j-6k

    where r is 3D vector and i,j,k are unit vectors in X,Y,Z directions.

    one thing about vectors is that number of dimensions is arbitrary. we can talks about 2, 3, 4, ... 10 or more dimensions.

    complex numbers can be seen as special case of vectors (just 2 dimensions) while real numbers are just special case of complex numbers (imaginary part is zero) or special case of vectors (all but one dimension are zero).

    if you like to relate it to 2D graph, real numbers are only along X axis (no unit vector is explicitly written) and imaginary componet is along Y axis. both values make one complex number.

    example of real world problem with more dimensionsions are area, volume etc.
    suppose two neighbours have pools. one is 10m long and the other is 8m.
    which one can hold more water? obviously length is not the only dimension that need to be factored in so by knowing only length, it is impossible to say which one has larger volume.

    similarily you cannot easily tell which of the two complex numbers or vectors has larger magnitude even if all components are mentioned. for example earlier mentioned r has magnitude:
    |r|=sqrt(5^2 + 3^2 + 6^2)
    |r|=8.3666

    if you have complex numbers such as:
    E1=2.4-2.6i
    or
    E2=2.5+2.5i
    which one is larger? it is not obvious but it is the first one because:
    |E1|=3.538361
    |E2|=3.535534

    you could use differrent representation where magnitude is obvious such as

    e=A*cos(a*t+p)

    so it is enough to compare magnitudes (A part) but many calculations quickly become quite difficult because you get several long terms with more than one sin or cos function.
     
    Last edited: Sep 5, 2011
  7. panic mode

    panic mode Well-Known Member

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  8. PG1995

    PG1995 Active Member

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    Thank you very much, everyone. You guys are really nice and helpful.

    Some may say mathematics is invented but I'm of the position that mathematics is discovered and development of most of mathematics has been driven by humans' struggle to mathematize the real life phenomena around them. We give interpretation to many of the concepts of mathematics. For instance, depending on the context we can interpret what a negative number mean. e.g. we understand what negative temperature mean, what negative velocity mean, what negative displacement mean. So, if we end up with a negative solution at the end in a certain problem then depending on the context we can figure out what that negative solution means. Now coming to the main point. Suppose, we end up with a complex solution, i.e., solution which contains complex number. What would it imply? Can you please give me some simple real life example which elaborate this? I hope you understand what I'm after.

    Regards
    PG
     
  9. Ratchit

    Ratchit Well-Known Member

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    PG1995 ,

    A complex quantity in the physical world means that a quantity has 2 or 3 parts. One part is designated as the reference, the second part is 90 degrees out of phase with the reference, and the third part if it exists is 90 degrees out of phase with the first and second parts. Complex power, voltage and current are good examples of 2 part (duplex) quantities.

    Ratch
     
  10. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Last edited: Sep 9, 2011
  11. misterT

    misterT Well-Known Member Most Helpful Member

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  12. Ratchit

    Ratchit Well-Known Member

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    misterT,

    Like many other descriptions of "j" or "i", I would fault it because it does not emphasize that j or i is a mathematical operator. They seem to treat it as a esoteric constant that does not have a well defined value in the real world, which is untrue. See my post #3 in this thread.

    Ratch
     
  13. misterT

    misterT Well-Known Member Most Helpful Member

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    i is not an operator, it is an imaginary number (constant) which has a real part of 0 and imaginary part of 1. Re(i)=0, Im(i)=1. Emphasizing i as an operator is a very big fault.

    How would you explain multiplying two operators, i*i = -1? Doesn't make sense.. unless i is not an operator, but a number.. a constant. And how about quaternions.. are there three operators? i, j and k?
     
    Last edited: Sep 10, 2011
  14. Ratchit

    Ratchit Well-Known Member

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    misterT,

    You lose on that argument. A lot of good math text books assert that it is an operator, just like ±,*,/,√, etc. You are beguiled into thinking it is a constant because you get correct results by treating it as a constant. As I explained earlier, it works that way because of its conformal similarity. Conceptionally 7j² means rotate 7 twice CCW by 90° to get -7. To avoid confusion, the designation should have been something like j(7) for 7j and j(j(7)) for 7j², but that is clumsy to write. So, as I mentioned before, one writes j for 1j.

    Sure it does. j*j means 1j*1j=1j², which means rotate 1 CCW by 90° twice or 180°, which equals -1. Look at the "Overview" section of this link http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/complex/transforms/index.htm

    Now, prove to me that you understand the concept. Find the value of j^j without using arithmetic or a calculator. Just use Euler's Identity and rotations to find the value.

    Ratch
     
  15. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi,
    As posted by Ratch, I was always taught that 'j' is an Operator, ie 90deg rotation, not a Constant..
     
  16. PG1995

    PG1995 Active Member

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    Many, many thanks for the replies. You guys are are very helpful.

    @Ratchit: I'm interested more in your explanation because it has at least given me an opportunity to think of imaginary numbers in terms of geometric interpretation. You say that 7j² (or, 7i²) means rotate 7 twice CCW by 90° to get -7. But what does simple 7j mean? Rotate by 90°? Please keep your exposition simple.

    I think what obstructs people like me from accepting j or i as an operator is the fact that it is not entirely like other operators such as + or x. You can say [Error: Syntax\ error : /usr/bin/latex --interaction=nonstopmode 2cf62586386934c63c008e55ee4a3d12.tex && /usr/bin/dvipng -q -D 300 -T tight -gamma 2.0 -bg Transparent -o 22473c7bd5e21db3e8392024f7e0124f-2.png 2cf62586386934c63c008e55ee4a3d12.dvi] but you can't speak of, let's say, addition operator as "+ = (something)".

    Thank you for the help.

    Regards
    PG
     
    Last edited: Sep 11, 2011
  17. Ratchit

    Ratchit Well-Known Member

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    PG1995,

    Yes, already explained in post #3 of this thread.

    It is wrong to think of operator j as a constant. Sometimes folks use j as an abreviation for 1/_90°, which is a constant. In that case, it is not being used as a operator anymore. You have to judge which is which by the context of the usage. So when used as a operator, j has no value. Neither do the other math operations ±,*,/,√,etc.

    Ratch
     
  18. misterT

    misterT Well-Known Member Most Helpful Member

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    I understand the concept of thinkin i as an operator, but it still is only a constant (imaginary unit). I can easily calculate with imaginary numbers without using i (or j). Using matrices for example.

    Here is an example of i*i. Because Re(i) = 0, and Im(i)=0, I get:
    Re(i*i) = 0*0 - 1*1 = -1
    Im(i*i) = 0*1 + 1*0 = 0

    I don't need the symbol i.. I only need to know the imaginary part and real part to do complex calculation. The constant i only makes using complex numbers easier.

    I can also write: i = e^(i*θ)
    where θ defines the complex argument, or "the amount of rotation", not i "as an operator of rotation". How does the operator concept fit into this form? How is it wrong to think the constant j as a constant (imaginary unit)? Can you prove it is not a constant?
     
    Last edited: Sep 11, 2011
  19. misterT

    misterT Well-Known Member Most Helpful Member

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    Well, j^j = e^(-π/2)
    But I don't see what this has to do with the "operator" concept. I would like to see you solve that only with Euler and rotations.
     
    Last edited: Sep 11, 2011
  20. Ratchit

    Ratchit Well-Known Member

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    misterT,

    No, i is not a constant when used as a operator. Many good textbooks say it is a operator.

    What does that prove? You are still using the concept of a rotational operator.

    Matrices are known to be able to perform rotations. Nothing new there.

    The operator i is not a constant when used as a operator. If you know what part of the number is orthogonal, you don't know need the operator i to tell you. It makes using duplex numbers easier because it tags the orthogonal number.

    No, i(1) = e^i(pi)/2 by Eulers Theorem.

    The exponent of e shows a rotation of pi/2 above.

    Because it is defined to be a operator, even if you can get correct results treating it like a constant.

    Yes, because it is defined to be a operator.

    OK, here is how it is done. i(1) = e^i(pi)/2 by Eulers Theorem. Notice the exponent of e is i(pi)/2. Applying the i operator to the exponent i(pi)/2 rotates it 90° CCW to -pi/2. That makes the term i(1)^i equal to e^-pi/2.

    You edited your post just in time, by the way.

    Ratch
     
  21. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again PG,

    In your original post, you said that sqrt(-1) was not allowed. But that's mainly only true in introductory algebra where they dont want to get into the added complexities of what sqrt(-1) might really mean.

    When you were younger and you learned the alphabet (assuming English) you learned the first three letters: A, B, and C. Now you might have went on to say that those where the only three letters 'allowed', but then you learned D, E, and F. You could have then again said that only those 6 letters were allowed, but then you went on to learn other letters of the alphabet.

    Well, other subjects are like this too. First you learn the basics, then you learn more advanced topics. Learning that sqrt(-1) is not allowed is a basic, but learning more about the meaning behind sqrt(-1) is a little more advanced, and like the alphabet, you just have to accept that there is more to learn.

    For the simplest example, you learned that the square root of a number is a number when multiplied by itself equals the original number. In other words, sqrt(x)*sqrt(x)=x. Well, if we do -1 we get sqrt(-1)*sqrt(-1)=-1 right? No problem there yet.

    Now when we begin to question what sqrt(-1) really is, we see that there is no "real" number that can satisfy this problem. However, there are other ways to deal with this kind of situation.

    One way is to simply never calculate it out. To leave it in that form and never do anything more with it. So an equation like this:
    y=2*sqrt(-1)+3+4

    simplifies to this:
    y=2*sqrt(-1)+7

    and we have 'solved' that equation, at least up to a point. To make it a little more readable we can introduce a variable 'j' to take the place of sqrt(-1) and we end up with this:
    y=2*j+7

    and this is as far as we can go. Note that we dont have to actually calculate the sqrt(-1) at any time here, but yet we still solved the equation.
    The end result of that we would say is a "complex number" because it contains two parts, one real and one we call "imaginary". The real part is 7 and the imaginary part is 2, and that's all we have to do. We are done. The specific real world interpretation of the real and imaginary parts will come from the application that we started with. For this example we didnt have a particular app, so this is pure mathematics so we're pretty much done.

    In the real world, we sometimes start with real numbers and end up with complex numbers, and sometimes we start with complex numbers and end up with real numbers, and sometimes we start with complex numbers and end with complex numbers. That's the way it is.


    To calculate X^Y with X and Y both complex we could use this:
    X^Y=e^(Y*ln(X))

    where in general the natural log and the exponentiation both have to be performed on complex numbers.
     
    Last edited: Sep 12, 2011

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