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Math Help - Complex Analysis

  1. #1
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    Complex Analysis

    Hi! Help me please with this one.

    Let's f(z) be an entire function
    and Re f(z)\geq 0for all z
    Prove that f is constant.

    Thanks.
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  2. #2
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    Re: Complex Analysis

    Quote Originally Posted by sinichko View Post
    Hi! Help me please with this one.

    Let's f(z) be an entire function
    and Re f(z)\geq 0 for all z
    Prove that f is constant.
    Hint: What can you say about the function e^{-f(z)} ?
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  3. #3
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    Re: Complex Analysis

    Quote Originally Posted by Opalg View Post
    Hint: What can you say about the function e^{-f(z)} ?
    I don't understand how it can help?
    e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}
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  4. #4
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    Re: Complex Analysis

    Quote Originally Posted by sinichko View Post
    I don't understand how it can help?
    e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}
    -f(z) = -u(x,y) - iv(x,y). So e^{-f(z)}=e^{-u(x,y)}(\cos v(x,y)-i\sin v(x,y)). You are told that u(x,y)\geqslant0 for all x,y. What does that tell you about |e^{-f(z)}| ?
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  5. #5
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    Re: Complex Analysis

    Quote Originally Posted by Opalg View Post
    -f(z) = -u(x,y) - iv(x,y). So e^{-f(z)}=e^{-u(x,y)}(\cos v(x,y)-i\sin v(x,y)). You are told that u(x,y)\geqslant0 for all x,y. What does that tell you about |e^{-f(z)}| ?

    |e^{-f(z)}| is bounded?
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  6. #6
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    Re: Complex Analysis

    Quote Originally Posted by sinichko View Post
    |e^{-f(z)}| is bounded?
    Yes! So what do you deduce from that?
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  7. #7
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    Re: Complex Analysis

    Quote Originally Posted by Opalg View Post
    Yes! So what do you deduce from that?
    Thanks.
    Maybe I should apply Liouville's theorem
    , but I have to prove that function e^{-f(z)}- an entire, given that f(z) - entire function...how can i prove that e^{-f(z)} entire function also?
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  8. #8
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    Re: Complex Analysis

    Quote Originally Posted by sinichko View Post
    Thanks.
    Maybe I should apply Liouville's theorem
    , but I have to prove that function e^{-f(z)}- an entire, given that f(z) - entire function...how can i prove that e^{-f(z)} entire function also?
    Yes, that is exactly what you need to do. In fact, a differentiable function of a differentiable function is always differentiable (that is basically what the chain rule says).
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