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Math Help - Need help solving this. Thank you.

  1. #1
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    Need help solving this. Thank you.

    Show that |e^(a+bi)|=e^a and Arg(e^(a+bi))=b.

    i know the first part.

    = (e^a)·(e^(bi))= (e^a) ( cos b + i sin b )
    =| e^(a+bi) | = (e^a)·√(cosē b + sinē b ) = e^a

    I cant get the second part
    Arg(e^(a+bi))=b
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  2. #2
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    Re: Need help solving this. Thank you.

    Quote Originally Posted by ilovemymath View Post
    Show that |e^(a+bi)|=e^a and Arg(e^(a+bi))=b.
    ...
    I cant get the second part Arg(e^(a+bi))=b
    That's because it's false if it's using the notation that I'm familiar with. It depends on whatever book you're using, but in the texts I've used, "Arg" with a capital A refers to the principle value of the polar angle - the one in between - \pi and \pi (I forget which endpoint is included - but only one of them is).

    (Counter) Example: Arg(e^{0 + 40.5 \pi i}) = Arg(e^{\frac{\pi}{2} i }) = \frac{\pi}{2}, so Arg(e^{0 + 40.5 \pi i}) \ne 40.5 \pi.
    Last edited by johnsomeone; September 25th 2012 at 02:15 PM.
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