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Math Help - Complex trig equation

  1. #1
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    Complex trig equation

    The question
    Find all solutions to  z \in C of:
    cosh(z) = 2i

    My attempt
    cosh(z) = \frac{e^z + e^{-z}}{2} = 2i

    e^z + e^{-z} = 4i

    e^{2z} + 1 = 4ie^z //multiply by e^z

    e^{2z} - 4ie^z = -1

    (e^z - 2i)^2 + 4 = -1 //complete the square

    e^z - 2i = \sqrt{-5}

    e^z = \sqrt{-5} + 2i

    e^z = i(\sqrt{5} + 2)

    e^xe^{iy} = (\sqrt{5} + 2)e^{\frac{\pi}{2}i}

    so e^x = \sqrt{5} + 2

    x = ln(\sqrt{5} + 2)

    y = \frac{\pi}{2} + 2k\pi //since exp is 2k\pi i periodic

    Since z = x + iy, z = ln(\sqrt{5} + 2) + i\pi(\frac{1}{2} + 2k)

    However, the solution is actually:
    (-1)^nsinh^{-1}(2) + (n + \frac{1}{2})\pi i

    I'm confused, my working looks correct. :/ Any help would be greatly appreciated!
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  2. #2
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    Re: Complex trig equation

    Quote Originally Posted by Glitch View Post
    The question
    Find all solutions to  z \in C of:
    cosh(z) = 2i

    My attempt
    \cosh(z) = \frac{e^z + e^{-z}}{2} = 2i

    e^z + e^{-z} = 4i

    e^{2z} + 1 = 4ie^z //multiply by e^z

    e^{2z} - 4ie^z = -1

    (e^z - 2i)^2 + 4 = -1 //complete the square

    e^z - 2i = \sqrt{-5} Should be \color{red}\pm\sqrt{-5}.

    e^z = \sqrt{-5} + 2i

    e^z = i(\sqrt{5} + 2)

    e^xe^{iy} = (\sqrt{5} + 2)e^{\frac{\pi}{2}i}

    so e^x = \sqrt{5} + 2

    x = \ln(\sqrt{5} + 2)

    y = \frac{\pi}{2} + 2k\pi //since exp is 2k\pi i -periodic

    Since z = x + iy, z = \ln(\sqrt{5} + 2) + i\pi(\frac{1}{2} + 2k)

    However, the solution is actually:
    (-1)^n\sinh^{-1}(2) + (n + \frac{1}{2})\pi i

    I'm confused, my working looks correct. :/ Any help would be greatly appreciated!
    Your answer is (half) correct, because \sinh^{-1}2 = \ln(\sqrt 5 + 2). See here, for example. You can get the other half of the answer by following through what happens if you use -\sqrt{-5} instead of \sqrt{-5}.
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  3. #3
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    Re: Complex trig equation

    You should know that \displaystyle \cosh{(z)} = \cosh{(x + iy)} = \cosh{(x)}\cos{(y)} + i\sinh{(x)}\sin{(y)}.

    So you are solving \displaystyle \cosh{(x)}\cos{(y)} + i\sinh{(x)}\sin{(y)} = 2i.

    Equating real and imaginary parts gives \displaystyle \cosh{(x)}\cos{(y)} = 0 and \displaystyle \sinh{(x)}\sin{(y)} = 2.

    Solving the first equation: Since \displaystyle \cosh{(x)} > 0 for all \displaystyle x, that means \displaystyle \cos{(y)} = 0 \implies y = \frac{(2n + 1)\pi}{2}, where \displaystyle n \in \mathbf{Z}.

    Substituting into the second equation \displaystyle \sinh{(x)}\sin{(y)} = 2 gives

    \displaystyle \begin{align*}\sinh{(x)}\sin{\left[\frac{(2n + 1)\pi}{2}\right]} &= 2 \\ (-1)^{n}\sinh{(x)} &= 2 \\ \sinh{(x)} &= \frac{2}{(-1)^n} \\ x &= \sinh^{-1}{\left[\frac{2}{(-1)^n}\right]} \end{align*}

    So the solution is \displaystyle z = \sinh^{-1}{\left[\frac{2}{(-1)^n}\right]} +  \frac{(2n + 1)\pi i}{2}, where \displaystyle n \in \mathbf{Z}.
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