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Homework Help: Complex Analysis - Solving Complex Trig functions

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Now, I know there's two ways to go about this and it seems everywhere I look around on the web people are solving it in a way I think that seems longer, harder and more prone to mistakes in exams. It involves using the exponential identities and taking logs. I was shown another way, but unfortunately I haven't quite got the grasp of it.

    sinh z = 0

    sinh (x + iy) = sinh(x)cos(y) + i cosh(x)sin(y) = 0


    1) sinh(x)cos(y) = 0
    2) cosh(x)sin(y) = 0

    (1) Either x = 0 or y = ± Pi/2
    (2) cosh (x) is never 0, so therefore x is not 0. Hence y = 0

    This is where I'm stuck, I do not know how to go from here.

    The answer is z = i*k*Pi
  2. jcsd
  3. Jun 13, 2012 #2
    You have two simultaneous equations. You've correctly deduced that cosh(x) is never 0, so therefore in equation 2 divide both sides by cosh(x) to get

    sin(y) = 0

    which occurs when y = k*Pi, where k is an element of the integer set

    Ok so we now know y! Let's look at x. Now you use your first equation:

    at y = k*Pi, cos(y) = 1 or -1

    so divide both sides by cos(y) and get

    sinh(x) = 0, therefore x = 0

    so what we have is z = x + iy = i * k * Pi.

    This method is a lot more simple, it just involves working through the simultaneous equations carefully. Where you started to go wrong was when you assumed y just has to equal 0, rather than the general solution of k * Pi
  4. Jun 13, 2012 #3
    Hmm, interesting. I think I'm starting to see it now.

    So suppose I have cosh z = 1

    cosh (x + iy) = cosh(x)cos(y) + i sinh(x)sin(y) = 1

    cosh(x)cos(y) = 1
    sinh(x)sin(y) = 0

    cosh(x) = 1 only when x = 0, and cos (y) = 1 for 2*k*Pi

    So now we already have values for x and y, substituting this into the imaginary component, validates that.

    So z = i*2*k*Pi

    Is my reasoning correct? Thank you for your help by the way.
  5. Jun 13, 2012 #4
    Stuck again now!

    cosh z = 2i

    I posted this question earlier but back then I only knew the exponential and log method.

    What I've got so far:

    cos (x + iy) = cosh(x)cos(y) + i sinh(x)sin(y) = 2i

    1) cosh(x)cos(y) = 0
    2) sinh(x)sin(y) = 2i

    From equation 1, cosh x cannot be zero. For cos (y) = 0, y is periodic so y = (Pi/2 + kPi)

    Now this is where I'm stuck on how to deal with that in equation 2.

    The final answer given is (-1)^k arcsinh(2) + i(kPi +Pi/2)

    I can see how that answer for y I got in equation 1 has something to do with the final answer.


    I'm thinking that if y = (Pi/2 + kPi), this would mean sin(y) = (-1)^k

    So dividing sin (y) in equation 2, we get sinh(x) = (-1)k 2i

    Taking arcsinh of this now means arcsinh((-1)^k 2i)

    Does this mean I can take out the (-1)^k ? I still haven't figured out what to do with the i hanging around.
    Last edited: Jun 13, 2012
  6. Jun 13, 2012 #5
    That's true but cosh(x) cos(y) = 1 holds for other values as well, for example cosh(x)=2, cos(y) = 1/2.

    that should be 2, not 2i
  7. Jun 13, 2012 #6
    That would explain it then, thanks!
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