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Math Help - Absolute Values of Complex Functions

  1. #1
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    Absolute Values of Complex Functions

    If f and g are continuous functions of a real variable and are non-zero in some interval, then clearly |f(x)| = |g(x)| implies f(x) = g(x) or f(x) = -g(x) on that interval.

    What is the analog for functions of a complex variable?

    This is my suspicion: if f and g are nonvanishing analytic complex-valued functions on some connected subset D of C, and |f(z)| = |g(z)| for all z in D, what can be said of f and g? Is it necessary that f = cg, where |c| = 1? If this is right, how would one prove it?

    Thanks in advance for any help.
    Last edited by joeyjoejoe; March 22nd 2010 at 05:21 PM.
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  2. #2
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    Quote Originally Posted by joeyjoejoe View Post
    If f and g are functions of a real variable and are non-zero in some interval, then clearly |f(x)| = |g(x)| implies f(x) = g(x) or f(x) = -g(x) on that interval.

    What is the analog for functions of a complex variable?

    This is my suspicion: if f and g are nonvanishing complex-valued functions on some connected subset of C, and |f(z)| = |g(z)| for all z in D, what can be said of f and g? Is it necessary that f = cg, where |c| = 1? If this is right, how would one prove it?

    Thanks in advance for any help.

    f(z)=e^{iRe(z)}\,,\,\,g(z)=1 . It is always true that |f(z)|=|g(z)| \,\,\forall z\in\mathbb{C} , but none is a scalar multiple of the other one, of course.

    Tonio
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  3. #3
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    Quote Originally Posted by joeyjoejoe View Post
    If f and g are functions of a real variable and are non-zero in some interval, then clearly |f(x)| = |g(x)| implies f(x) = g(x) or f(x) = -g(x) on that interval.
    What if  g(x) = \begin{cases} f(x), & \mbox{if }  x \geq 0  \\ -f(x),  & \mbox{if } x < 0. \end{cases} .

    Here we have  |f(x)|=|g(x)| but  f(x)\neq\pm g(x) .
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  4. #4
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    I meant to say f and g are continuous (real variable case) and analytic (complex variable case) respectively.

    I don't think that counterexample you give is analytic, so I'm still thinking it could be true.
    Last edited by joeyjoejoe; March 22nd 2010 at 05:16 PM.
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