Let $U\subseteq \mathbb{C}$ be an open set, $z_{0}\in U$ and $R>0$ such that $\mathcal{B}_{R}(z_{0}) \subseteq U$. Let $f:U\rightarrow \mathbb{C}$ be a holomorphic function with Taylor's serie $f(z)=\sum_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ where $R$ is the Radius of convergence.
Show that for all $n\in \mathbb{N}_{>0}$ and $0<r<R$ $$c_{n}r^{n}=\frac{1}{2\pi}\int_{0}^{2\pi}f\left(z_{0}+re^{-2nt}\right)dt$$
Remark: I am for thinking that there is a mistake in the statement of the problem because I have tried but have not succeeded.
My attempt: Considering the curve $$\begin{array}{rll} \gamma : [0,2\pi)& \longrightarrow & \mathbb{C} \\ t& \longmapsto & z_{0}+re^{it} \end{array} $$ We have $$ \begin{array}{rcl} c_{n}r^{n} &=& \frac{r^{n}}{2 \pi i}\intop_{0}^{2 \pi} \frac{f\left(z_{0}+re^{it}\right)}{\left(z_{0}+re^{it}-z_{0}\right)^{n+1}}rie^{it} dt \\ &=& \frac{1}{2 \pi }\intop_{0}^{2 \pi} f\left(z_{0}+re^{it}\right)e^{-nit} dt.\\ \end{array} $$ To conclude the prove would be sufficient to show that $$f\left(z_{0}+re^{it}\right)e^{-nit}=f\left(z_{0}+re^{-2nt}\right).$$ But I could not show this. I honestly think it's false but I could not find an contra-example.
Question: I need to know if the problem is well posed, if this ill-posed please give me an example. If this well posed please tell me who fails in my attempt and tell me how I can finish the prove.