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

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

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

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

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

    This follows from the calculation

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

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

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

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

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