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Math Help - complex exponentials

  1. #1
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    complex exponentials

    May be its not the right place to ask this, but mods can move this to other parts of the forum(i encountered it while leaning diff equations)...

    I would like to know how to find \Re(-\frac{2}{e^{ix}+i}) in trigonometric form or polar form, so please try to show all the steps.

    It would also be very helpful if someone could post a link to learn converting complex exponentials to trignometric functions, or just how to find out the real part of the complex exponential in trignometric form.

    Thanks
    Last edited by earthboy; January 26th 2013 at 01:48 AM.
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  2. #2
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    Re: complex exponentials

    You need to make use of Euler's formula: \displaystyle \begin{align*} e^{i\,\theta} = \cos{(\theta)} + i\sin{(\theta)} \end{align*}.
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    Re: complex exponentials

    Quote Originally Posted by Prove It View Post
    You need to make use of Euler's formula: \displaystyle \begin{align*} e^{i\,\theta} = \cos{(\theta)} + i\sin{(\theta)} \end{align*}.
    I know, but how do i do it???
    please solve it fully or post a link where similar problems are solved...
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    Re: complex exponentials

    Quote Originally Posted by earthboy View Post
    I know, but how do i do it???
    please solve it fully or post a link where similar problems are solved...
    I will do neither. YOU will try this. When you write down the exponential in terms of its real and imaginary parts using Euler's formula, you will have a complex fraction. You should know that you simplify fractions by multiplying top and bottom by the conjugate.
    Thanks from earthboy
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    Re: complex exponentials

    Quote Originally Posted by Prove It View Post
    I will do neither. YOU will try this. When you write down the exponential in terms of its real and imaginary parts using Euler's formula, you will have a complex fraction. You should know that you simplify fractions by multiplying top and bottom by the conjugate.
    \frac{-2}{e^{ix}+i},which by applying euler's formula becomes \frac{-2}{\cos x+i \sin x +i}=\frac{-2}{\cos x+i(\sin x+1)}
    multiplying it by conjugate becomes \frac{-2(\cos x-i(\sin x+1))}{(\cos x+i(\sin x+1))(\cos x-i(\sin x+1)}
    as we are dealing with the real part only it becomes \frac{-2\cos x}{2\sin x +2}=-\frac{\cos x}{\sin x+1}
    right???
    Thanks for the hint though....and to euler too..
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    Re: complex exponentials

    Yes that is correct
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