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Square root of -1 .

  1. Jun 17, 2005 #1
    Square root of -1.....

    We say that the square root of -1 is equal to i ( or j ), and that this is therefore not a real number - but what is this fact actually useful for? Why over complecate things with something not on the number line - why is it so useful to treat the square root of -1 as a number? :tongue2:
    Thanks in advance. :smile:
     
  2. jcsd
  3. Jun 17, 2005 #2
    Because it is one, mathematics in the complex plane is a branch of its own and has proven to be important even in the physical sense. The two branches of numbers are real and imaginary, but both are equally numbers.
     
  4. Jun 17, 2005 #3

    HallsofIvy

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    For one thing it simplifies enormously sin(x) and cos(x) so that they can be written as exponentials. That allows engineers to use the much simpler exponential function to represent waves rather than sine and cosine.
     
  5. Jun 18, 2005 #4

    dextercioby

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    Even though the introduction of complex (imaginary) numbers in mathematics happened when 3 italian fellows were studying the 3-rd order algebraic equation with integer coefficients,i'm sure they were really happy to say that the equation

    [tex] x^{2}+1 =0[/tex] has solutions.Two of them.

    And then claim that every possible polynomial of 2-nd,3-rd & 4-th degree has roots.

    Daniel.
     
  6. Jun 18, 2005 #5

    uart

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    Cheman, one simple answer to your question is to point out that for many mathematical problems that involve only real inputs and only real outputs that the simplist solution method can involve complex numbers at intermediate steps.
     
  7. Jun 18, 2005 #6

    Hurkyl

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    Another interesting fact for those who prefer to live in the reals...

    There exist real numbers that can be expressed through the application of arithmetic and n-th roots to integers, but there absolutely, positively must be complex intermediate values to do so.
     
  8. Jun 19, 2005 #7
    Could you elaborate or provide a link? I'm curious.
     
  9. Jun 19, 2005 #8

    Gokul43201

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    Complex numbers are useful for a whole lot of stuff and essential for most of the following - here's a few randomly selected choices :

    1. AC electrical theory
    2. Vibration analysis
    3. Number Theory (to study something even as "real" as the distribution of the prime numbers)
    4. Quantum Mechanics, Field Theories and Relativity
    5. Digital Signal Processing
     
  10. Jun 19, 2005 #9

    Hurkyl

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    Unfortunately, I can't. I think I first heard the fact when we were going over Galois theory in my algebra class, but I can't remember exactly how it worked.
     
  11. Jun 19, 2005 #10
    It worked well for me in ac circuits...

    When iw as at school and i asked my teacher about it, he told me it was practically usefull for engineers in everything, mostly for prediction..
     
  12. Jun 19, 2005 #11
    Nobody saw fit to mention, that one of the very first reasons "i" was useful is because of this kind of mistake:

    [tex]\sqrt-1*\sqrt-1=\sqrt{-1*-1}=\sqrt1[/tex]
     
    Last edited: Jun 19, 2005
  13. Jun 19, 2005 #12

    Hurkyl

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    The mistake there is that √(-1) is not defined in the reals. So, if you're living in real number land, using √(-1) in an arithmetic expression is exactly as meaningless as doing division by zero.
     
  14. Jun 19, 2005 #13
    HurkyL: Unfortunately, I can't. I think I first heard the fact when we were going over Galois theory in my algebra class, but I can't remember exactly how it worked.

    Well, what about this? Define [tex]\omega=\frac{-1+\sqrt-3}{2}[/tex], a cube root of 1, and let [tex]\omega'[/tex] represent it conjugate. Then define Y as:

    Y = [tex](\omega)^\frac{1}{3} +(\omega')^\frac{1}{3}[/tex]

    If we cube this value we get the equation Y^3-3Y+1=0. An equation with three real roots.
     
    Last edited: Jun 19, 2005
  15. Jun 19, 2005 #14

    Hurkyl

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    The problem is that I don't remember how to prove that there isn't some other method of constructing Y that does not involve complexes.
     
  16. Jun 19, 2005 #15
    Hurkyl: The problem is that I don't remember how to prove that there isn't some other method of constructing Y that does not involve complexes.


    Evidently not, from what I am reading: "Well, there: you've got an answer using the tools of Galois theory and solvability. Note that there isn't a single reference to the reals or complexes. They are not really part of Galois theory, although they can be incorporated into the discussion if one wishes. Most of us usually don't wish to do so; I take it you do.

    Very well then, your question seems to be, "Is there always a sequence of fields F_i _all of which are contained in the real numbers_ which have the properties (1)-(3) stated above? (That's not what "solvability" means in Galois theory, though!) To this question, the answer is no -- in fact, never, for irreducible cubics. Obviously condition (2) would require all three roots to be real, but that is the so-called "casus irreducibilis" which seems to prompt your query. A cubic with three real roots which factors in F_{i+1} would also factor in F_i. This is an exercise in Garling's Galois Theory, p. 138.

    I'm not sure what the point of this is. It is easily proved that the roots of (say) x^3-3x-1 are all real, and can be expressed in terms of radicals like (1/2 + i sqrt(3)/2)^(1/3) + (1/2 - i sqrt(3)/2)^(1/3). This is an exact expression involving complex numbers whose imaginary part is clearly exactly zero. It may be manipulated like any other complex number. The result I quoted you above shows that its real part cannot be expressed using a finite number of radical operations applied only to real numbers; so? Sounds to me like a good reason to use the complex form -- this way we have something to write down. If you don't like this form, you may use the trigonometric forms; I hardly think these are "better" but of course may be better suited to some particular purposes. Chacun a son gout.

    Dr. Rusin calls the above solution (for one of the three roots) a solution in "complex radicals" and declares that it qualifies as "a solution in radicals" according to the Galois theory. Such a solution would not be accepted here or in any other shop that seeks practical solutions to practical problems. In this case the solution given cannot be expressed in terms of real radicals because the given equation is an irreducible cubic according to the definition given on the previous page. The root can be expressed in trigonometric terms, namely 2cos(π/9). This is an exact expression because it can be evaluated to any desired degree of accuracy." http://www25.brinkster.com/ranmath/misund/poly02.htm
     
    Last edited: Jun 19, 2005
  17. Jun 19, 2005 #16

    Hurkyl

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    As I recall, this fact was one of the major motivations for accepting the complexes in standard mathematics. Mathematicians of the time were happy enough to say that quadratics that had no real solutions simply didn't have a solution -- they saw no point in accepting the complexes simply so they could say all quadratics had a root.

    But the cubic and quartic formulas changed that -- even if you only cared about real roots, you still needed to take a tour through the complexes to get there.

    Historically speaking, this result was significant.
     
  18. Jun 19, 2005 #17
    On a more general level, what exactly makes the set of complex numbers more useful than the set of reals? Is it the fact that they're algebraically closed?
     
  19. Jun 19, 2005 #18
    What exactly does the "hypercomplex" numbers not satisfy that the reals and complex do?
     
  20. Jun 20, 2005 #19

    lurflurf

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    What exactly does the "hypercomplex" numbers not satisfy that the reals and complex do?
    The main thing they lack is the hypercomplex nombers are noncommutative.

    Being Algebraically closed is quite nice.
    x^2+1=0 has solutions.
    Outside of that poles at complex points often help explain "mysterious" behaviour on the real line. Like why the Taylor series of 1/(1+x^2) about 0 diverges if |x|>0. Many integrals, products, and sums are easy to compute using residue theory. Inverting integral transforms such as those of Laplace is sometimes easier using the complex inversion formula than the real one. It is often nice (as has been mentioned) to allow intermidiate steps or small errors to involve complex numbers. Many formulas look nice using complex numbers. Like (cos(x)+i sin(x))^n=cos(n x)+i sin(n x) for example. Some problems in R^2 are nice when expressed in C^1. Many time a method is general with complex numbers, but several cases must be dealt with over reals. For example consider the differential equation y''+1=0 if complex variables are used the solution is in the form C1 exp(r1 x)+C2 exp(r2 x) where r1,r2 are roots of r^2+1=0. This is similar to what happens with y''-1=0, but if complex numbers are not used we have to treat y''+1=0 as a special case and use sine and cosine. Also many other equations can be solved like sin(x)=17 and exp(x)+1=0.
     
  21. Jun 20, 2005 #20

    lurflurf

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    That quoted page has a small error it interchanges x^3-3x+1=0 and x^3-3x-1=0.
    Clearly as they are odd functions the roots of one are the additive inverses of roots of the other.
    i.e.
    2cos(pi/9) is a root of x^3-3x-1
    -2cos(pi/9) is a root of x^3-3x+1
    Also care must be taken as the roots obtained by using particular cube roots do not correspond.
     
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