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Homework Help: Cos(x)=-2 DEFINED?

  1. Jul 27, 2010 #1
    1. The problem statement, all variables and given/known data

    I was trying to solve that problem ignoring what people say about it not being defined and got a number strangely. I showed step by step so you can tell me what I did wrong as I don't think I did anything wrong what so ever.



    2. Relevant equations

    In work below

    3. The attempt at a solution

    e^(ix) = cosx + i sinx = cisx
    e^(-ix) = cosx - i sinx = cis(-x)
    cis(x) + cis(-x) = cosx + i sin x + cosx - i sinx = 2cosx
    from which
    cosx = (cis(x) + cis(-x) )/2
    from which I set it equal to -2 and began to solve
    cosx = (cis(x) + cis(-x) )/2 = -2
    Multipled by 2 on both sides
    cis(x) + cis(-x) ) = -4
    Multipled by cisx on both sides
    cis^2(x) + 1 = -4cis(x)
    set equal to zero by adding -4cis(x) to both sides
    cis^2(x) + 4cis(x) + 1 = 0
    solved the quadratic for cis(x)
    (-4 +/- sqrt(4^2-4(1)))/2
    = -2 +/- sqrt(16 - 4)/2
    = -2 +/- sqrt(12)/2
    = -2 +/- (2sqrt(3))/2
    = -2 +/- sqrt(3)
    cis(x) = -2 +/- sqrt(3) = e^(ix)
    solved for x took natural log of both sides
    ln(-2 +/- sqrt(3)) = ln( e^(ix) )
    took out exponents and used ln(e) = 1
    ln(-2 +/- sqrt(3)) = ix
    divided through by i
    ln(-2 +/- sqrt(3)) /i = x
    simpified for +/-
    for +
    ln(-2 + sqrt(3)) /i = x
    for -
    ln(-2 - sqrt(3)) /i
    factored out negative one
    ln(-(2 + sqrt(3))/i
    used the fact that ln(xy) = lnx + lny
    ( ln(-1) + ln(2 + sqrt(3)) )/i
    used the fact that ln(-1) = i pi
    ( i pi + ln(2 + sqrt(3)) )/i
    canceled out the i
    pi + ln((2 + sqrt(3)) )/i = x
     
  2. jcsd
  3. Jul 27, 2010 #2

    eumyang

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    (Post removed. Completely misunderstood the question.)
     
    Last edited: Jul 27, 2010
  4. Jul 27, 2010 #3

    CompuChip

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    I wouldn't say you did anything wrong, rather, you are talking about a different function.
    People who tell you that cos(x) = -2 has no solution, are talking about the function from [0, 2pi] to [-1, 1] and then they are right.
    If you consider the function defined from C to C (in general) by 1/2(cis(x) + cis(-x)) then you will indeed obtain the answer you got.

    [edit] Actually I can kinda agree with you. The definition of the exponential on the complex plane can be rigorously given (in several ways, just avoid the one with sin(x) and cos(x) of course :D) and then defining cos(x) and sin(x) as being the restrictions of (exp(ix) +/- exp(-ix))/2 to the real number is rather reasonable.
     
    Last edited: Jul 27, 2010
  5. Jul 27, 2010 #4
    Actually I don't know I just have been told... so I treid it for myself and got a number... so therefore it does exist no?
     
  6. Jul 27, 2010 #5
    What do you mean from C to C what is C?
     
  7. Jul 27, 2010 #6

    eumyang

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    C = set of complex numbers
     
  8. Jul 27, 2010 #7

    tiny-tim

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    Hi GreenPrint! :smile:

    (have a pi: π and a square-root: √ and try using the X2 icon just above the Reply box :wink:)

    Starting with your final answer, which is cos(π + ln(2 + √3)/i),

    (and since cos(-A) = cosA, and we can always add any whole multiple of 2π, and since1/i = -i, the general solution would be: cos((2n+1)π ± i ln(2 + √3)),

    we can immediately use cos(π + A) = cosπcosA - sinπsinA = -cosA, to get:

    [STRIKE]cos(i ln(2 + √3)),[/STRIKE]

    -cos(-i ln(2 + √3)).

    ok, let's check that answer, using cosx = (eix + e-ix)/2:

    [STRIKE]cos(i ln(2 + √3)) = (e-ln(2 + √3) + eln(2 + √3))/2

    = (1/(2 + √3) + (2 + √3))/2 = ((2 - √3) + (2 + √3))/2 = 2.

    hmm :confused: … something seems to have gone wrong. :redface:[/STRIKE]

    -cos(i ln(2 + √3)) = -(e-ln(2 + √3) + eln(2 + √3))/2

    = -(1/(2 + √3) + (2 + √3))/2 = -((2 - √3) + (2 + √3))/2 = -2. :biggrin:
     
    Last edited: Jul 27, 2010
  9. Jul 27, 2010 #8
    So if I'm not specifically told we are in the set of real numbers or it's not implied any where or have never been told in class that we will always be in the set of real numbers then indeed I had to solve for cosx = -2 ??? So am I justified in solving for it because it's a summer assignment and so I've never been told to always assume we are in the set of real numbers so I should leave that as a solution on my paper and put "Assuming we are also in set of complex of numbers.." or something?
     
  10. Jul 27, 2010 #9

    Mentallic

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    C means in the complex numbers. When people say it's not defined, they mean it's not defined in the real numbers. In the same way one asks where does the graph y=x2+1 and the x-axis cut each other, well, not at any real number, but it does in the complex numbers.

    Now, just to get a bit picky, the log function in the complex plane is multi-valued, so the answer is actually [tex]x=2\pi n +\pi +ln(2+\sqrt{3})/i[/tex]

    But you can do better than this. Can you manipulate the above to be in the form x=a+ib?
    Rather than dividing by i, it's more conventional to multiply by i.
     
  11. Jul 27, 2010 #10
    hmmm let me look at that... also is that n symbol another symbol for pi or is it suppose to be pi just wounder I'd rather use n for pi then the other symbol...
     
  12. Jul 27, 2010 #11

    Mentallic

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    No no no! He had the right answer, you multiplied by i when he divided by i! :tongue:
    Oh and it wouldn't work for n=2,4,6... rather the formula is [tex]x=\pi(2n+1)+ln(2+\sqrt{3})/i[/tex]

    Sorry to be so harsh hehe
     
  13. Jul 27, 2010 #12
    hmmm interesting your right... so just multiply threw by i? lets see...
     
  14. Jul 27, 2010 #13
    pi i (2n + 1) + ln(2 + sqrt(3) )
    ?
     
  15. Jul 27, 2010 #14

    Mentallic

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    I delved into the complex numbers back in high school when we were learning about quadratics. My teacher got really pissy at me :biggrin:
    But ever since I would just act like a smart-xxx on my assignments and get into way more detail than required, just for the laughs. So be my guest!
     
  16. Jul 27, 2010 #15

    Mentallic

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    Nope, what I mean is to multiple the fraction [tex]\frac{ln(2+\sqrt{3})}{i}[/tex] by i in both the numerator and denominator so nothing changes. The denominator will become a real number :wink:
     
  17. Jul 27, 2010 #16
    well that would give i^2 in the denomenator or simply -1 and i in the numerator

    -i ln(2+ sqrt(3))
    ?
    don't see how that is a real number hmmm
     
  18. Jul 27, 2010 #17
    By definition, as you previously did:

    [tex]
    \cos(z) \equiv \frac{e^{i \, z} + e^{-i \, z}}{2}, \; z \in \mathbb{C}
    [/tex]

    So, one can try and find the inverse function [itex]w = \arccos(z)[/itex] by solving with respect to [itex]w[/itex]:

    [tex]
    z = \cos(w) = \frac{e^{i \, w} + e^{-i \, w}}{2}
    [/tex]

    [tex]
    e^{2 \, i \, w} - 2 \, z \, e^{i \, w} + 1 = 0
    [/tex]

    This is a quadratic equation with respect to [itex]e^{i \, w}[/itex], with a solution:

    [tex]
    e^{i \, w} = z + (z^{2} - 1)^{1/2}
    [/tex]

    where the square root has two branches. Taking the logarithm (an infinitely valued complex function):

    [tex]
    w = \arccos(z) \equiv = -i \, \log{\left(z + (z^{2} - 1)^{1/2}\right)}
    [/tex]
     
  19. Jul 27, 2010 #18
    ok so your also saying that
    pi (2n + 1) - i ln(2 + sqrt(3) ) = - i ln (2 + sqrt(3) ) I don't see how?
     
  20. Jul 27, 2010 #19
    I don't understand what you are trying to do here? :confused:
     
  21. Jul 27, 2010 #20

    Mentallic

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    Hang on, let's stop at [tex]x=\pi(2n+1)-iln(2+\sqrt{3})[/tex]

    This is correct, now don't change anything drastically. Just use the rules you know. This can be simplified a bit further. Use the log rule that [tex]-ln(x)=ln(x^{-1})[/tex] and then rationalize the denominator inside the log function. It's not changing the answer, just making things look a little more neat I suppose :smile:
     
  22. Jul 27, 2010 #21
    If you use the definition, you will have:

    [tex]
    -i \, \log{(-2 + \sqrt{3})}
    [/tex]

    [tex]
    -i \, log{(-2 - \sqrt{3})}
    [/tex]

    Then, you need the definition of the complex logarithm. It is the inverse function of the complex exponential Let [itex]z = \rho \, e^{i \, \theta}[/itex] and [itex]w = u + i \, v[/itex] and [itex]z = \exp(w)[/itex]. Using the properties of the exponential function:

    [tex]
    \rho \, e^{i \, \theta} = e^{u} \, e^{i v}
    [/tex]

    This means that:
    [tex]
    \rho = e^{u} \Rightarrow u = \ln{\rho}, \; \rho > 0
    [/tex]

    [tex]
    v = \theta + 2 \, n \, \pi, \; n \in \mathbb{Z}
    [/tex]

    So, a complex logarithm is a multiple valued function (with infinite number of branches):

    [tex]
    w = \log{z} \equiv = \ln{\rho} + i \, \left(\theta + 2 \, n \, \pi), \; z = \rho \, e^{i \, \theta}, \; \rho > 0, \; n \in \mathbb{Z}
    [/tex]
     
  23. Jul 27, 2010 #22
    ok then

    pi(2n + 1) + i ln(2 - sqrt(3) )
    hmmm
     
  24. Jul 27, 2010 #23
    I get a different solution with the imaginary part with the opposite sign. Also, there is another branch of the square root that you hadn't considered.
     
  25. Jul 27, 2010 #24
    I don't understand why
    pi(2n + 1) + i ln(2 - sqrt(3) ) = i ln(2 - sqrt(3) )
    i get that it has branches but
    pi(2(5) + 2) + i ln(2 - sqrt(3) ) does not equal i ln(2 - sqrt(3) )
     
  26. Jul 27, 2010 #25
    If you look at my definition of the arc cosine, you will see that your right hand sides are not in accordance with it.
     
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