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Thread: Complex trig equation

  1. #1
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    Complex trig equation

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

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

    $\displaystyle e^z + e^{-z} = 4i$

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

    $\displaystyle e^{2z} - 4ie^z = -1$

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

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

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

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

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

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

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

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

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

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

    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$\displaystyle z \in C$ of:
    cosh(z) = 2i

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

    $\displaystyle e^z + e^{-z} = 4i$

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

    $\displaystyle e^{2z} - 4ie^z = -1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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