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    I remember doing something similar for FP2 but that was a while ago now and I'm a little stuck with:

    Find all the complex roots of

    \sinh{z} = 1

    So far:

    I have gotten up to

    e^z = 1 \pm \sqrt{2}

    but I am unsure of how to get the other lot of roots except for the obvious.
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    If you expand and suchlike, you get e^z - e^{-z} = 2... from there, if you let z = x + iy, then:
    \newline

e^z - e^{-z} = 2\newline

\Rightarrow e^x(\cos y + i\sin y) - e^{-x}(\cos (-y) + i\sin (-y)) = 2\newline

\Rightarrow e^x(\cos y + i\sin y) - e^{-x}(\cos y - i\sin y) = 2\newline

\Rightarrow (e^x - e^{-x})\cos y + (e^x + e^{-x})i\sin y = 2 + 0i\newline

\Rightarrow \begin{cases} (e^x - e^{-x})\cos y = 2\\ (e^x + e^{-x})\sin y = 0\end{cases}

    Then it's just simultaneous equations.

    Spoiler:
    Show
    Note that e^x + e^{-x} > 0 for all x, so \sin y = 0
    Spoiler:
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    \newline y = k\pi
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    (Original post by nuodai)
    If you expand and suchlike, you get e^z - e^{-z} = 2... from there, if you let z = x + iy, then:
    \newline

e^z - e^{-z} = 2\newline

\Rightarrow e^x(\cos y + i\sin y) - e^{-x}(\cos (-y) + i\sin (-y)) = 2\newline

\Rightarrow e^x(\cos y + i\sin y) - e^{-x}(\cos y - i\sin y) = 2\newline

\Rightarrow (e^x - e^{-x})\cos y + (e^x + e^{-x})i\sin y = 2 + 0i\newline

\Rightarrow \begin{cases} (e^x - e^{-x})\cos y = 2\\ (e^x + e^{-x})\sin y = 0\end{cases}

    Then it's just simultaneous equations.

    Spoiler:
    Show
    Note that e^x + e^{-x} > 0 for all x, so \sin y = 0
    Spoiler:
    Show
    \newline y = k\pi
    So checking I'm on the right lines:

    e^x + e^{-x} \not=0 \Rightarrow \sin(y)=0

\newline \Rightarrow y=2\pi n 

\newline \Rightarrow \cos(y) = \pm 1 

\newline \Rightarrow e^x - e^{-x} = \pm 2 

\newline \rightarrow x = \ln( \pm 1+\sqrt{2}) 

\newline \newline \Rightarrow z=\ln(\sqrt{2}\pm 1) + \pi n i

    Would this be the right answer?

    Edit: I didn't notice the spoiler and have just noticed I missed out half of the solutions - will update in a minute when I worked it though with  n\pi

    Edit: updated
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    (Original post by The Muon)
    So checking I'm on the right lines:

    e^x + e^{-x} \not=0 \Rightarrow \sin(y)=0 \newline \Rightarrow y=2\pi n \newline \Rightarrow \cos(y) = 1 \newline \Rightarrow e^x - e^{-x} = 2 \newline \rightarrow x = \ln(1+\sqrt{2}) \newline \newline \Rightarrow z=\ln(1+\sqrt{2}) + 2\pi n i

    Would this be the right answer?
    Well y = k\pi, not just 2k\pi; when k is even, \cos y = 1, so \ln (1 + \sqrt{2}) + k\pi i is right; but when k is odd, \cos y = -1, so e^{x} - e^{-x} = -2 and so on.
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    Just updated my post.
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    (Original post by The Muon)
    Just updated my post.
    Looks about right to me Except the 2\pi n i should still just be \pi n i (unless n is multiples of a half).
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    (Original post by nuodai)
    Looks about right to me Except the 2\pi n i should still just be \pi n i (unless n is multiples of a half).
    Yeah, I just forgot to edit it out. I'd rep you if it wasn't for the stupid 28 day rule :mad:

    And expect a couple more question in this thread - I have a few problems and I'm still rusty
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    Can you have a quick glance through this and point out my mistakes :p:

    Find all roots of:

     \cos(z)=\sqrt{2}

    My solution:

     \cos(z) = \cosh(iz) = \sqrt{2}

\newline \Rightarrow e^{iz} +e^{-iz} = 2\sqrt{2}

\newline \Rightarrow e^{-y}( \cos(x) + i \sin(x) ) + e^{y}( \cos(x) - i \sin(x) )



\newline \Rightarrow \begin{cases} (e^{-y} + e^{y})\cos(x) = 2\sqrt{2} \\ (e^{-y} - e^{y})\sin (x) = 0\end{cases}



\newline \Rightarrow sin(x)=0 \text{ trying the other part = 0 doesn't work out }

\newline \Rightarrow x = \pi n

\newline \Rightarrow e^{y} +e^{-y} = \pm 2\sqrt{2}

\newline \Rightarrow e^{2y} \pm 2\sqrt{2} e^{y} + 1 = 0

\newline \Rightarrow e^{y} = \pm \sqrt{2} \pm 1

\newline \Rightarrow y= \ln(\sqrt{2} \pm 1) 



\text{ and then just putting it into the form} z = x+iy \text{ for the final answer}
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    ln of a negative number causes a lot of problem.
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    (Original post by Simplicity)
    ln of a negative number causes a lot of problem.
    Cheers simp. I always knew you'd pull through and get my silly mistake at least once.

    Is that you in your profile pic?
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    (Original post by The Muon)
    Cheers simp. I always knew you'd pull through and get my silly mistake at least once.

    Is that you in your profile pic?
    Well, you can have ln of a negative number but you really need to write it in this form. ln(1- \sqrt 2) + i \pi n but that has a ton of problems like having infinite amount of solutions.

    Yes.
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    (Original post by The Muon)
    Is that you in your profile pic?
    How does one upload a profile picture if you don't mind me asking?
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    (Original post by DeanK22)
    How does one upload a profile picture if you don't mind me asking?
    You go to my account (top right) then click on the word images and profile pic is there.
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    (Original post by The Muon)
    You go to my account (top right) then click on the word images and profile pic is there.
    Aah - embarrassingly simple. Merci.
 
 
 
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