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Thread: Complex Analysis

  1. #1
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    Complex Analysis

    Hi! Help me please with this one.

    Let's $\displaystyle f(z)$ be an entire function
    and $\displaystyle Re f(z)\geq 0$for all $\displaystyle z$
    Prove that $\displaystyle f$ 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 $\displaystyle f(z)$ be an entire function
    and $\displaystyle Re f(z)\geq 0$ for all $\displaystyle z$
    Prove that $\displaystyle f$ is constant.
    Hint: What can you say about the function $\displaystyle e^{-f(z)}$ ?
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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 $\displaystyle e^{-f(z)}$ ?
    I don't understand how it can help?
    $\displaystyle e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}$
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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?
    $\displaystyle e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}$
    $\displaystyle -f(z) = -u(x,y) - iv(x,y)$. So $\displaystyle e^{-f(z)}=e^{-u(x,y)}(\cos v(x,y)-i\sin v(x,y))$. You are told that $\displaystyle u(x,y)\geqslant0$ for all x,y. What does that tell you about $\displaystyle |e^{-f(z)}|$ ?
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  5. #5
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    Re: Complex Analysis

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

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

    Quote Originally Posted by sinichko View Post
    $\displaystyle |e^{-f(z)}|$ 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 $\displaystyle e^{-f(z)}$- an entire, given that $\displaystyle f(z)$ - entire function...how can i prove that $\displaystyle e^{-f(z)}$ 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 $\displaystyle e^{-f(z)}$- an entire, given that $\displaystyle f(z)$ - entire function...how can i prove that $\displaystyle e^{-f(z)}$ 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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