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Math Help - Find a complex number

  1. #1
    MHF Contributor arbolis's Avatar
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    Find a complex number

    I'd like to check my result.
    Find all the complex numbers z such that e^{i \overline z}= \overline {e^{iz}}.
    I've found that z=x, where x \in \mathbb{R}. So when z is any real number.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I'd like to check my result.
    Find all the complex numbers z such that e^{i \overline z}= \overline {e^{iz}}.
    I've found that z=x, where x \in \mathbb{R}. So when z is any real number.
    Could you show your work?

    Also, consider z=\pi/2. Then e^{i\overline{\pi/2}}=e^{i\pi/2}=i\neq -i = \overline{e^{i \pi/2}}
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  3. #3
    Super Member Random Variable's Avatar
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    EDIT. I made a mistake.

     e^{i\overline{z}} = e^{i(x-iy)}=e^{y+ix} = e^{y}e^{ix}= e^{y}\big(\cos(x)+i\sin(x)\big)

     \overline{e^{iz}} = \overline{e^{i(x+iy)}} = \overline{e^{-y+ ix}} = \overline{e^{-y}e^{ix}} =  e^{-y}\big( \cos(x) - i \sin(x)\big)

    so it seems to be for only  z= n \pi + 0i \ \ n= 0, \pm 1, \pm 2, ...
    Last edited by Random Variable; August 22nd 2009 at 04:14 PM.
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  4. #4
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    Quote Originally Posted by arbolis View Post
    I'd like to check my result.
    Find all the complex numbers z such that e^{i \overline z}= \overline {e^{iz}}.
    I've found that z=x, where x \in \mathbb{R}. So when z is any real number.
    No. If z= x, a real number, i\overline{z}= iz while \overline{iz}= -iz In general, e^{iz}\ne e^{-iz}.
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  5. #5
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    Quote Originally Posted by Random Variable View Post
    EDIT. I made a mistake.

     e^{i\overline{z}} = e^{i(x-iy)}=e^{y+ix} = e^{y}e^{ix}= e^{y}\big(\cos(x)+i\sin(x)\big)

     \overline{e^{iz}} = \overline{e^{i(x+iy)}} = \overline{e^{-y+ ix}} = \overline{e^{-y}e^{ix}} =  e^{-y}\big( \cos(x) - i \sin(x)\big)

    so it seems to be for only  z= n \pi + 0i \ \ n= 0, \pm 1, \pm 2, ...
    Okay, if z= n\pi, then e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n while e^{\overline{i n\pi}}= e^{-i n\pi}= cos(-n\pi)+ i sin(n\pi)= (-1)^n since cosine is an even function.
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Okay, if z= n\pi, then e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n while e^{\overline{i n\pi}}= e^{-i n\pi}= cos(-n\pi)+ i sin(n\pi)= (-1)^n since cosine is an even function.
    So are you saying I'm right?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Okay, if z= n\pi, then e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n while e^{\overline{i n\pi}}= e^{-i n\pi}= cos(-n\pi)+ i sin(n\pi)= (-1)^n since cosine is an even function.
    Actually it is if z= n\pi, then e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n while \overline{e^{i n\pi}}= \overline{cos(n\pi)+ i sin(n\pi)} = cos(n\pi) - i sin(n\pi) = (-1)^n

    because in the right hand it is \overline{e^{iz}} and not e^{\overline{iz}}

    even if it is the same since

    \cos x - i \sin x)" alt="\overline{e^{iz}} = \overline{e^{i(x+iy)}} = \overline{e^{-y}e^{ix}} = e^{-y}\\cos x - i \sin x)" /> while

    \cos x - i \sin x)" alt="e^{\overline{iz}} = e^{-i \overline{z}} = e^{-i (x-iy)} = e^{-y}\\cos x - i \sin x)" />
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