If $\displaystyle f$ is an entire function and maps $\displaystyle D(0,1)$ onto itself then it maps $\displaystyle \bar D(0,1)$ onto itself.

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- May 8th 2008, 11:59 AMThePerfectHackerEntire Function on a Disk
If $\displaystyle f$ is an entire function and maps $\displaystyle D(0,1)$ onto itself then it maps $\displaystyle \bar D(0,1)$ onto itself.