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Imaginary Numbers

  1. Aug 15, 2004 #1

    Alkatran

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    I've been doing a lot of thinking about imaginary numbers lately. My first question was "What is sqr(i)?".

    I thought it was unsolvable until I punched it into my trusty (and often right) caclulator and found out it was (sqr(.5)i + sqr(.5))^2
    So obvious now. Of course.

    Anyways, a while later, I thought about series of powers (the same way a power is a series of multiplications and those a series of additions). (We'll call a series of power ^^)

    So my question is:
    x^^2 = x^x = -4
    What is x?

    My calculator tells me: false. But I have this tendency not to trust it sometimes, especially when it won't give me an answer.

    Also, is i ^ i defined? ((-1)^.5)^((-1)^.5) = (-1)^(i/2) = ?
     
  2. jcsd
  3. Aug 15, 2004 #2

    robphy

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    Use the identity [itex]e^{i\theta}=\cos\theta + i\sin\theta [/itex]. Observe that [itex]e^{i(\pi/2)}=\cos(\pi/2) + i\sin(\pi/2)=i[/itex]

    So, [itex]i^i=(e^{i\pi/2})^i=e^{i^2\pi/2}=e^{-\pi/2}\approx 0.2078795764[/itex]
     
  4. Aug 15, 2004 #3

    Alkatran

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    Thanks
    (I haven't... learned that identity yet :redface: )
     
  5. Aug 15, 2004 #4
    I think it was Cardan who got concerned with such problems: Find two numbers such that their sum is 4 and their product is 13.
     
  6. Aug 15, 2004 #5

    Alkatran

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    x+y = 4
    x=4-y
    x*y=13
    x=13/y
    13/y=4-y
    13=4y-y^2
    -y^2+4y-13 = 0
    y^2-4y+13=0
    y = (4 +- sqr(16 - 4*1*13))/2 = 2 +- sqr(16 - 52) = 2+- sqr(-36) = 2+-6i

    Oh, your point was that he discovered imaginary numbers?
     
  7. Aug 15, 2004 #6
    Well, Cardan seemed to have, now and then, suggested that imaginary numbers might have some actual mathematical worth, which was not the usual attitude.
     
  8. Aug 15, 2004 #7

    mathwonk

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    alkatran,

    the number e^x is always equal to 1+x + x^2/2 + x^3/6 + x^4/24 +...+x^n/n!+...

    and whenever you want to evaluate a^b, the easiest way is to write a as e^(lan(a)),
    so that a^b = [e^ln(a)]^b = e^[ln(a).b] = e ^b.ln(a). For example, since e^(r+si), with r,s real, equals (e^r)[cos(s)+isin(s)], and sin(pi/2) = 1, then e^(i pi/2) = i, so ln(i) may be taken to be i pi/2, although there are many other values of z such that e^z = i.

    Anyway, one value of ln(i) is i pi/2, so i^i can be taken as e^i[ln(i)] = e^i [i pi/2]
    = e^[- pi/2].

    To see why e^it = cos(t) + i sin(t), work out the Taylor series

    e^(it) = 1 + (it) + [(it)^2]/2 + [(it)^3]/6 + ....+ [(it^n]/n!+....


    amd cos(t) = 1 - t^2/2 + t^4/4! -+.....


    and sin(t) = t - t^3/3! + t^5/5!-+.........

    and add them....


    Assuming you have had Taylor series.

    Or use the fact that any soution of f '' + f = 0, and f(0) = a and f'(0) = b, must equal acos(t) + bsin(t).

    I.e. then check that f(t) = e^(it) solves f '' + f =0, and also f(0) = 1, and f '(0) = i.

    hence f(t) = e^(it) must equal cos(t) + i sin(t).
     
  9. Aug 15, 2004 #8

    Alkatran

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    Great post mathwonk, just so you know, I graduated from high school two months ago.

    We learned that (if I remember correctly):
    x^n = cos(nO) + i*sin(nO)
    where x is a complexe number and O (theta) is the angle from (0,0) to your complexe number on a graph.

    Actually, I'm positive that formula is wrong (in this post, not yours!). There's more to it.


    The explanation you gave is excellent but is going to take some time to digest.
     
  10. Aug 16, 2004 #9

    mathwonk

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    Actually your formula might be right for complex numbers of length one. For any arrow in the plane, such as a complex number, making angle O with the origin, the x coordinate is equal to rcos(O), where r is the length, and the y coordinate is equal to rsin(O). Hence it is true for any complex number x (not zero), that x = rcos(O) + i rsin(O). Now to raise a complex number to a power means to raise the length to that power, and multiply the angle by n,

    so x^n would be r^n cos(nO) + i r^n sin(nO).

    so yes you are right your formula is missing the r^n factor.

    (I myself got this wrong also while writing the original version of this post a few minutes ago.)
     
  11. Aug 16, 2004 #10

    mathwonk

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    As to exponentials of complex numbers, they are simpler when the base is a positive real number a, since then there is a unique natural choice of logarithm, the unique real number b = ln(a) such that e^b = a.

    Then we can define a^c = e^cb.

    But if the base is not a positive real number, then there is not so natural a choice for the log, so we have to choose between infinitely many of them, as for i. Still it makes no difference which one we choose, we get the same value of a^c.
     
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