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    Area of the image of an annulus under a univalent Laurent series

    E-TDDQI

    Area
    Complex Analysis
    Topics
    Conformal Maps, Integrals, Laurent Series
    Status
    draft

    Let \(f(z) = \sum_{n= -\infty}^\infty c_n z^n\) be analytic and one-to-one in \(r_0< |z| < R_0\). For \(r_0<r<R<R_0\), let \(D(r,R)\) be the annulus \(r<|z|<R\). Show that the area of \(f(D(r,R))\) is finite and is given by \[S = \pi \sum_{n=- \infty}^\infty n |c_n|^2 (R^{2n} - r^{2n}).\]

    Solution

    See above solution: all goes identically up until the integral over \(r\) values, just replace \(\int_0^R\) with \(\int_r^R\).