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Homework Help: Comlex Logarithms Question

  1. Aug 14, 2010 #1
    1. The problem statement, all variables and given/known data

    When was trying to solve this problem
    cos(x)=-2 I got the following answers
    i ln(2 +/- sqrt(3)) + 2 pi n
    n set of integers positive and negative
    now I was told by a couple of people on here to add 2 pi n to my solution because before it didn't have it. At the time I didn't understand why but now I do. All I did was simply think about it as an angle and hence adding multiples of 2 pi n give valid solutions but what I don't understand is why does
    http://www.wolframalpha.com/input/?i=(i^((2x)/pi)+i^((-2x)/pi))/(2)=-2
    give me the solutions of
    x = 2 pi n + pi - i log(2 +/- sqrt(3) ), n set of positive and negative integers ?
    Now please understand I get that there are an infinite number of solutions. I completely understand this. This is the reason why when we define a function as a mathematical formula or representation that only has one output value for each input value of it's domain is false, correct? The complex logarithm is a completley valid function. So I understand there are an infinite amount of solutions I just don't understand were that other pi came from or the -i. I was woundering if somebody cuold please explain this to me. Thank You!

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 14, 2010 #2
    anybody please help?
     
  4. Aug 14, 2010 #3

    Char. Limit

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    You got your answer wrong. The inverse of cos(y)=x is

    [tex]y=\frac{log\left(x\pm\sqrt{x^2-1}\right)}{i}[/tex]

    Can you spot your mistakes? There are two.
     
  5. Aug 14, 2010 #4
    Iw asn't dong the inverse of cosine though sorry I forgot to mention it I was doing this

    cos(x)=-2

    so you don't have that domain restriction correct?
     
  6. Aug 14, 2010 #5
    and I have no idea were that equation came from... also it's not normal to divide by i right
     
  7. Aug 14, 2010 #6

    Char. Limit

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    I actually messed it up and was fixing it when you posted.

    The domain restriction doesn't apply here, and dividing by i is just multiplying by -i... it's necessary.
     
  8. Aug 14, 2010 #7
    ok well unforunatley i don't know were the mistakes are =(...
     
  9. Aug 14, 2010 #8

    Char. Limit

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    First, you multiplied by i instead of -i.

    Second, the 2 in your answer should be a -2.

    After a bit of playing with logarithmic properties, I managed to turn one of the answers into W-A's answer.
     
  10. Aug 14, 2010 #9
    Tahnk youu
     
  11. Aug 15, 2010 #10
    still need help

    Ok well wolfram's answer was this
    x = 2 pi n + pi -i log(2-sqrt(3)), n element Z
    and
    x = 2 pi n + pi -i log(2+sqrt(3)), n element Z

    now when i solved it I got this for my answers
    x = iln(2+3^.5) + 2 pi n
    and
    iln(2-3^.5) + 2 pi n
    n set of integers positive and negative

    the two is not negative and I still don't know where that other pi came from in wolfram's answer can you please help
     
  12. Aug 16, 2010 #11

    Char. Limit

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    Well, let's start with my answer, given by my inverse formula for cos(x)...

    [tex]i\left(-log\left(-2\pm\sqrt{3}\right)\right)[/tex]

    Now if you factor the -1 out of the inside and then use the rule [itex]alog(b)=log(b^a)[/itex], you get this:

    [tex]ilog\left(\frac{-1}{2\pm\sqrt{3}}\right)[/tex]

    Which after using the rule [itex]log\left(\frac{a}{b}\right)=log(a)-log(b)[/itex] becomes...

    [tex]ilog(-1)-ilog(2\pm\sqrt{3})[/tex]

    From there it's just recognizing that [itex]ilog(-1)=-\pi=\pi+2\pi n[/itex] for n=-1, an integer.
     
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