1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    So

    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.

    Edit:

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex Analysis - Solving Complex Trig functions
Loading...