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Math Help - Limit question

  1. #1
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    Limit question

    Suppose f is Complex differentiable everywhere and that Lim|z|→∞}f(z)=1
    Prove f(z) is constant
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  2. #2
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    You have to use Liouville theorem. Find a closed ball B such that outside the modulus of f is smaller than 2, and apply that the closed ball is compact to conclude that f is bounded-
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  3. #3
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    Quote Originally Posted by thespian View Post
    Suppose f is Complex differentiable everywhere and that Lim|z|→∞}f(z)=1
    Prove f(z) is constant

    First proof: Liouville's theorem states that a holomorphic (analytic) function which is bounded must be constant, and that's exactly what happens with our function: in some circle |f(z)|<K\,,\;\;\forall\;|z|<R\,,\,\,and\;\;|f(z)|<  1.5\;\forall |z|>R,because \;f(z)\xrightarrow [|z| \to \infty] {} 1

    Second proof: By Cauchy's integral representation and by the estimation lemma: |f'(z)|=\left|\frac{1}{2\pi i}\oint\limits_{|z|=R}\frac{f(w)\;dw}{(w-z)^2}\right|\leq \frac{1}{2\pi} 2\pi R \frac{\max\limits_{|z|=R} (|f(z)|)}{R^2}=\frac{\max\limits_{|z|=R} (|f(z)|)}{R} \xrightarrow [R \to \infty] {} 0, so f'(z)=0 \Longrightarrow f(z) constant.

    Tonio
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