Results 1 to 6 of 6

Math Help - Complex Analysis Help Please (not complicated :D)

  1. #1
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132

    Complex Analysis Help Please (not complicated :D)

    I'm not 100% sure of the solution to the following problem:

    Find all the solutions of:


    , where z=x+iy


    I attempted doing this question the way my lecture notes seem to indicate, as follows:

    ,







    , where

    But i'm really not sure this makes any sense?! Can anyone help me out of confusion? Is there another way of doing this or is this vaugely right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,910
    Thanks
    1760
    Awards
    1
    Quote Originally Posted by Jason Bourne View Post
    I'm not 100% sure of the solution to the following problem: Find all the solutions of:
    , where z=x+iy
    The standard way is to write cos(z) in the u+iv form.
    \cos (z) = \cos (x)\cosh (y) + i\left( { - \sin (x)\sinh (y)} \right).
    So we get \cos (x)\cosh (y) = 0\;\& \; - \sin (x)\sinh (y) = 3.
    Now note that \cosh (y) \ne 0\; \Rightarrow \;\cos (x) = 0\; \Rightarrow \;x = \frac{{\left( {2k + 1} \right)\pi }}{2}.
    Now you can finish?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by Jason Bourne View Post
    I'm not 100% sure of the solution to the following problem:

    Find all the solutions of:


    , where z=x+iy


    I attempted doing this question the way my lecture notes seem to indicate, as follows:

    ,







    , where

    But i'm really not sure this makes any sense?! Can anyone help me out of confusion? Is there another way of doing this or is this vaugely right?
    Please don't double post. See rule #1 here.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Jason Bourne View Post
    Find all the solutions of:

    , where z=x+iy

    ...

    , where
    This solution is completely correct. You could simplify it a bit by using properties of the logarithm: \ln(i(3\pm\sqrt10)) = \ln i + \ln(3\pm\sqrt10), and \ln i = \ln(e^{i\pi/2}) = i\pi/2. So z = 2n\pi - i\ln(i(3\pm\sqrt10)) = (2n+{\textstyle \frac12})\pi-i\ln(3\pm\sqrt10).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132
    Thankyou all for your help and sorry for the double post. I'm not confused now but was wondering, the real part suggested by Plato is <br />
x = \frac{{\left( {2n + 1} \right)\pi }}{2}<br />

    but the real part calculated using the other method is <br />
x = (2n+{\textstyle \frac12})\pi
    am I missing something here? Thanks Anyway.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Jason Bourne View Post
    the real part suggested by Plato is <br />
x = \frac{{\left( {2n + 1} \right)\pi }}{2}<br />

    but the real part calculated using the other method is <br />
x = (2n+{\textstyle \frac12})\pi
    am I missing something here?
    That's a very interesting question. The answer is that there is no discrepancy between the two solutions, though it took me a while to figure out how they fit together.

    Start with the solution in the form z = (2n+{\textstyle \frac12})\pi-i\ln(3\pm\sqrt10). There are two solutions for each value of n, namely

    z = (2n+{\textstyle \frac12})\pi-i\ln(3+\sqrt10) [1] and z = (2n+{\textstyle \frac12})\pi-i\ln(3-\sqrt10) [2].

    Looking at the second of these, the first thing you should notice is that 3√10 is negative. In fact, 3-\sqrt{10} = \frac{-1}{3+\sqrt{10}} (this follows from the fact that (3+\sqrt{10})(3-\sqrt{10}) = 9-10=-1). Therefore \ln(3-\sqrt{10}) = \ln(-1) - \ln(3+\sqrt{10}) = i\pi - \ln(3+\sqrt{10}), and we can rewrite [2] as z = (2n+1+{\textstyle \frac12})\pi+i\ln(3+\sqrt10). This can be combined with [1] to form the single expression z = (n+{\textstyle \frac12})\pi-i(-1)^n\ln(3+\sqrt10), which agrees with Plato's method for finding the solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 06:30 AM
  2. Replies: 6
    Last Post: September 13th 2011, 08:16 AM
  3. Replies: 1
    Last Post: October 2nd 2010, 02:54 PM
  4. Replies: 12
    Last Post: June 2nd 2010, 03:30 PM
  5. Replies: 1
    Last Post: March 3rd 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum