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Math Help - tanh(2+3i)

  1. #1
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    tanh(2+3i)

    \displaystyle tanh(2+3\mathbf{i})=\frac{\mathbf{i}sinh(2+3\mathb  f{i})}{cosh(2+3\mathbf{i})}

    \displaystyle \frac{cos(3)sinh(2)+\mathbf{i}sin(3)cosh(2)}{cos(3  )cosh(2)+\mathbf{i}sin(3)sinh(2)}

    How can I simplify this further without multiplying by the conjugate?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    \displaystyle tanh(2+3\mathbf{i})=\frac{\mathbf{i}sinh(2+3\mathb  f{i})}{cosh(2+3\mathbf{i})}

    \displaystyle \frac{cos(3)sinh(2)+\mathbf{i}sin(3)cosh(2)}{cos(3  )cosh(2)+\mathbf{i}sin(3)sinh(2)}

    How can I simplify this further without multiplying by the conjugate?
    Hint: Use the sum formula for tangent and recall that \text{tanh}(x)=-i\tan(ix).
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  3. #3
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    Also, \displaystyle \tanh{(2+3i)} = \frac{\sinh{(2+3i)}}{\cosh{(2+3i)}}, not \displaystyle \frac{i\sinh{(2+3i)}}{\cosh{(2+3i)}}.


    Anyway...

    \displaystyle \tanh{(2+3i)} = \frac{e^{2(2+3i)}-1}{e^{2(2+3i)}+1}

    \displaystyle = 1 - \frac{2}{e^{2(2+3i)}-1}

    \displaystyle = 1 - \frac{2}{e^{4 + 6i}-1}

    \displaystyle = 1 - \frac{2}{e^4\,e^{6i} - 1}

    \displaystyle = 1 - \frac{2}{e^4(\cos{6} + i\sin{6}) - 1}

    \displaystyle = 1 - \frac{2}{e^4\cos{6}-1 + i\,e^4\sin{6}}

    \displaystyle = 1 - \left(\frac{2}{e^4\cos{6} - 1 + i\,e^4\sin{6}}\right)\left(\frac{e^4\cos{6} - 1 - i\,e^4\sin{6}}{e^4\cos{6} - 1 - i\,e^4\sin{6}}\right)

    \displaystyle = 1 - \frac{2(e^4\cos{6} - 1 - i\,e^4\sin{6})}{(e^4\cos{6} - 1)^2 + (e^4\sin{6})^2}

    \displaystyle = 1 - \frac{2e^4\cos{6} - 2 - 2i\,e^4\sin{6}}{e^8\cos^2{6} - 2e^4\cos{6} + 1 + e^8\sin^2{6}}

    \displaystyle = 1 - \frac{2e^4\cos{6} - 2 - 2i\,e^4\sin{6}}{e^8 - 2e^4\cos{6} + 1}

    \displaystyle = \frac{e^8 - 2e^4\cos{6} + 1 - (2e^4\cos{6} - 2 - 2i\,e^4\sin{6})}{e^8 - 2e^4\cos{6} + 1}

    \displaystyle = \frac{e^8 - 4e^4\cos{6} + 3 + 2i\,e^4\sin{6}}{e^8 - 2e^4\cos{6} + 1}

    \displaystyle = \frac{e^8 - 4e^4\cos{6} + 3}{e^8 - 2e^4\cos{6} + 1} + \left(\frac{2e^4\sin{6}}{e^8-2e^4\cos{6} + 1}\right)i
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  4. #4
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    \displaystyle tan(\mathbf{i}z)=\frac{\mathbf{i}sinh(z)}{cosh(z)}
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