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Math Help - Locally injective holomorphic map

  1. #1
    Member HappyJoe's Avatar
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    Locally injective holomorphic map

    Hello,

    I'm having trouble with part of the following problem:

    Suppose G is an open set in the complex plane \mathbb{C}, and let a be a point of G. Let f be a holomorphic function on G, and suppose f'(a)\neq 0.

    First I'm suppose to show that there exists an r>0, such that |f'(z)-f'(a)|<|f'(a)| for all z\in B_r(a), the ball of radius r and center a.

    This follows from continuity of f' at a, with epsilon =|f'(a)|.

    Then I'm showing that for all z_1,z_2\in B_r(a), it holds that

    |\int_0^1 f'(tz_2+(1-t)z_1)\ dt - f'(a)| < |f'(a)|.

    This follows from the calculation

    |\int_0^1 f'(tz_2+(1-t)z_1)\ dt - f'(a)|
    = |\int_0^1 f'(tz_2+(1-t)z_1) - f'(a)\ dt|
    \leq  \int_0^1 |f'(tz_2+(1-t)z_1) - f'(a)|\ dt
    < \int_0^1 |f'(a)|\ dt = |f'(a)|,

    with the last inequality following from the fact that the line segment between z_1 and z_2 is contained in B_r(a).

    Then I'm asked to prove that for all z_1 and z_2 in B_r(a), it is true that
    f(z_2)-f(z_1) = (z_2-z_1)\int_0^1 f'(tz_2+(1-t)z_1)\ dt,

    which I do by substituting a single variable for the expression tz_2+(1-t)z_1 in the integral.

    My problem is to show that this implies that f restricted to the ball B_r(a) is injective, which boils down to showing that the integral is non-zero for z_1\neq z_2. Any ideas?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Well, if 0=\int_0^1 f'(tz_2+(1-t)z_1)\ dt, then |\int_0^1 f'(tz_2+(1-t)z_1)\ dt - f'(a)| < |f'(a)| implies |f'(a)|<|f'(a)|...
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  3. #3
    Member HappyJoe's Avatar
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    Ah! I missed this, thanks for pointing it out.
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