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Math Help - Roots of equation

  1. #1
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    Roots of equation

    Problem statement is to find all roots of cos(z) - isin(z) = 0

    I know that e^(-iz) = cos(z) - isin(z)

    If I were to express e^(-iz) = e^y*e^(-ix) = 0
    (since z = x + iy)
    either e^y = 0 or e^(-ix) = 0 or both.
    e^(-ix) = cos(x) - isin(x). This is never 0.
    e^y as a real function of y is also never 0.

    Thus, how do I go about finding the roots to this equation?
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  2. #2
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    Quote Originally Posted by amoeba View Post
    Problem statement is to find all roots of cos(z) - isin(z) = 0

    I know that e^(-iz) = cos(z) - isin(z)

    If I were to express e^(-iz) = e^y*e^(-ix) = 0
    (since z = x + iy)
    either e^y = 0 or e^(-ix) = 0 or both.
    e^(-ix) = cos(x) - isin(x). This is never 0.
    e^y as a real function of y is also never 0.

    Thus, how do I go about finding the roots to this equation?

    As \cos z=\frac{e^{iz}+e^{-iz}}{2}\,,\,\,\sin z=\frac{e^{iz}-e^{-iz}}{2i}\Longrightarrow\,0=\cos z-i\sin z=e^{-iz}

    Putting z=x+iy\,,\,\,x\,,\,y\,\in\mathbb{R}, we get 0=e^{-iz}=e^{-ix+y}=e^y(\cos x-i\sin x)\Longrightarrow\,\cos x=\sin x= 0, and since these two real functions never vanish at the same point we get that there are no roots.

    Tonio
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