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Thread: exponential

  1. #1
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    exponential

    how do you solve l 1- e^(ix) l = k l sin (x/2) l for all real x.
    fink k.

    thanks!
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    how do you solve l 1- e^(ix) l = k l sin (x/2) l for all real x.
    fink k.

    thanks!
    $\displaystyle |1-e^{ix}|=k\left|\sin \frac{x}{2}\right| \Longrightarrow |1-\cos x-i\sin x|^2=k^2\sin^2\frac{x}{2}\Longrightarrow$ $\displaystyle 1-2\cos x+1=k^2\sin^2\frac{x}{2}\Longrightarrow 4\sin^2\frac{x}{2}=k^2\sin^2\frac{x}{2}$

    $\displaystyle \Longrightarrow\,\, either\,\,\sin \frac{x}{2}=0\Longleftrightarrow \frac{x}{2}=2m\pi\,,\,\,m\in \mathbb{Z} \,,\,\,or\,\,\,k^2=4\Longleftrightarrow k=\pm 2$

    We used above $\displaystyle \cos x =\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$ , and also $\displaystyle \forall\,x\,,\,y\in \mathbb{R}\,,\,\,|x+iy|^2=x^2+y^2$

    Tonio
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  3. #3
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    by the way, how do you get from the 2nd equation to the 3rd equation of (1-2cos x +1)?
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