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Surjectivity of Sin/Cos over the Complex Plane

  1. Feb 5, 2007 #1
    Verify Identities Cos(z) = (exp(iz)+exp(-iz))/2 sin(z)= (exp(iz)-exp(-iz))/2i

    Using identities show that sin(z) and cos(z) are surjective from C to C.

    Determine all z in C such that sin(z) = 12i/5

    Solutions: The verification seems simple using eulers formula and the even/odd nature of cos/sin. My problem is using the identities to show sin/cos are surjective.

    Given m in C, m = cos(z) = (exp(iz)+exp(-iz))/2 implying 2m = exp(iz)+exp(-iz) implying Log(2m) = Log(exp(iz)(1+exp(-2iz)) = izLog(1+exp(-2iz). From here I was going to expand the log into a taylor series.. though I'm not sure how that will help. Any suggestions would be appreciated. Obviously, once I can find some equation to determine z, I can just plug it in to find z for 12i/5 and then add steps of 2ipi since Log is periodic over the complex numbers.

    Just spent around 2 hours trying to tackle the problem from a different perspective... not sure how to proceed. Any help, if just a possible hint from someone who is unsure themselves would be appreciated.
    Last edited: Feb 6, 2007
  2. jcsd
  3. Feb 6, 2007 #2
    Alright, I've made some progress. Given m in C, m = cos(z) =
    1/2[e^iz + e^-iz] implying 2m = e^-iz[1+e^2iz] implying
    2me^iz = 1 + e^2iz implying e^2iz -2me^iz +1 = 0 implying
    e^iz = 2m +or- Sqrt(4m^2 - 4) / 2 = m+or- Sqrt(m^2 - 1)
    Thus: z = -iLog[m +or- (m^2-1)^(1/2)]mod2pi

    Problem: When I did the same thing for sin(z) I got the same answer... Is that correct? Furthermore, the last part of the problem is to find all z such that m = 12i/5.

    sin(z) = 12i/5 what are possible values of z? I'm a little unsure how to proceed... or at least give my answer.

    I could just leave it as z = -iLog(12i/5 +or- iSqrt(119/25)) Though I assume there is a way to simplify it more.
  4. Feb 6, 2007 #3


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    What you are saying is that you have absolutely no idea what the problem is asking!

    Apparently you are trying to solve for values of v that make cos(z)= (1/2)(eiz- eiz) true. The problem asked you to verify that it is an identity- that is, that it is true for all z.

    Start by writing out Taylor series for cos(z), eiz, and e-iz.
  5. Feb 6, 2007 #4


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    No... He says he understands the identities. What he is trying to do is construct an explicit inverse for sin and cos (ie given m=sin(z) find a value for z). He wants to show they are surjective.
    Last edited: Feb 6, 2007
  6. Feb 6, 2007 #5


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    I think you are essentially getting it. You wind up with quadratic equations for e^(iz) and then you finally get z with a log. Be careful though, you aren't going to get only two solutions for z. exp(i*z*(1+2*pi*n))=exp(i*z) for all integers n.
  7. Feb 6, 2007 #6


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    And you don't get the same equation for sin and cos. You are just being careless.
  8. Feb 6, 2007 #7
    Thx for checking it Dick, its good to know if I made a mistake :P. It felt funny that the same thing came out for sin that happened in cosine. I'll go back and redo the quadratic and see what went wrong.

    Edit: Found problem, I divided by 2i instead of 2 when solving the quadratic.
    Last edited: Feb 6, 2007
  9. Feb 6, 2007 #8
    Ok, redid the sin(z) and I got a different answer.
    m = sin(z) = 1/2[e^iz - e^-iz]
    2im = e^iz - e^-iz = e^-iz(e^2iz - 1)
    2ime^iz = e^2iz - 1
    e^2iz - 2ime^iz - 1 = 0
    e^iz = [2im +/- (-4m^2 + 4)^(1/2)]/2
    e^iz = im +/- i(m^2 - 1)^(1/2)]
    iz = Log[im +/- i(m^2 - 1)^(1/2)] mod2pi
    z = -iLog[im +/- i(m^2 - 1)^(1/2)] mod2pi (Since its going in sin its periodic)

    So, if m = 12i/5 then:

    z = -iLog(-12/5 +/- i(-199/25 - 1)^(1/2))mod2pi
    z = -iLog[-12/5 +/- -(198)^(1/2)/5]mod2pi

    once you get these answers you can add or subtract steps of 2ipiK for all k in Z since log/the exponential are periodice by 2ipi.

    Is there a way to easily simplify this Log statement so that the values are a little more concrete?

    Using QuickMath Simplification... doesnt seem like you can take the log all that well. All it did was turn it into the form:

    z = pi - iLog[1/5(12+(199)^(1/2)]
    pi coming from Log(-1), taking 25 out of the sqroot and then taking 1/5 out of the expression. I gave my answer as above and then added steps of 2ipi for an infinite set of solutions. Thx for your help! If you find any mistakes reading this dont hestitate to remark!
    Last edited: Feb 6, 2007
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