On an entire function

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

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

Solution

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

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

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

\begin{align*} \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{align*}

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