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On an entire function

Let n \in \mathbb{N} and f be an entire function. Prove that for any arbitrary positive numbers a, b it holds that:

\displaystyle \frac{\bigintsss_{0}^{2\pi} e^{-i n t}f \left ( z+a e^{it} \right ) \, {\rm d}t}{\bigintsss_{0}^{2\pi} e^{-i n t} f\left ( z + b e^{it} \right ) \, {\rm d}t} = \left ( \frac{a}{b} \right )^n


Since our function is entire this means that it is holomorphic and can be represented in the form

\displaystyle f(x)=\sum_{m=0}^{\infty} a_m \left ( x-z \right )^m

This series converges uniformly on [0, 2\pi] thus we can interchange summation and integral. Hence:

\begin{aligned} \int_{0}^{2\pi} e^{-in t} f\left ( z + ae^{it} \right ) \, {\rm d}t &= \int_{0}^{2\pi} \sum_{m=0}^{\infty} a_m a^m e^{it \left ( m-n \right )} \, {\rm d}t \\ &= \sum_{m=0}^{\infty} a_m a^m \int_{0}^{2\pi}e^{it \left ( m-n \right )} \, {\rm d}t \\ &= 2 \pi \sum_{m=0}^{\infty} a_m a^m \delta_{mn}\\ &=2 \pi a_n a^n \end{aligned}

where \delta_{mn} is Kronecker’s delta. Similarly for the denominator. Dividing we get the result.

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