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Suppose that f is a one-to-one analytic function mapping the disc $|z|<1$ onto a bounded domain D. Show that the area of D is given by $$A(D)=\pi \sum_{n=1}^{\infty} n|a_n|^2$$, where $\sum_{n=0}^{\infty} a_n z^n$ is the power series for f in $|z|<1$.

This is a practice problem from Fisher's complex variables chapter 3.5 and I can use the fact that $$A(D)=\int \int_{|z|<1} {|f'(z)|}^2 dxdy$$ for the above problem. I am not sure where to start but I am considering using the summation of singularities and cauchy integral to get the result (although I am not sure how to apply them.)

james black
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  • Differentiate term by term and use polar coordinates to evaluate the double integral.. – Kavi Rama Murthy Jan 25 '21 at 23:18
  • and then it would equal to the double integral of |f'(z)|^2? if so, i am not sure how to convert it to polar coordinates, can you show it for just one term thanks – james black Jan 25 '21 at 23:23
  • Use $dxdy=rd\theta dr$ and $|f'(z)|^2=f'(z)\bar f'(z)$ multiply term by term and use orthogonality to get rid of the cross terms when you integrate $d\theta$ so remain with only diagonal terms that are functions of $r$ only; technically to switch series and integrals you need uniform convergence so one does it first on the disc of radius $\rho$ and then let $\rho \to 1$ – Conrad Jan 26 '21 at 00:05

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