The following is part of an exercise I am demonstrating, below I will ask my question.
I will detail a little of what I have:
Let $f$ be a holomorphic function on $\mathbb{C}$, $f$ is injective. Let's define $\gamma(t):=e^{it}$, for $t\in[0,2\pi]$ and $\beta(t):=\gamma(-t)=e^{-it} $, for $t\in[0,2\pi]$.
Then by Cauchy we have that $f(z)=\sum_{n=0}^{\infty}a_nz^n$ where $a_n:=\int_{\gamma}\frac{f(w)}{w^{n+1}}dw$.
Let's define $g(z):=f(1/z)$. Note that $g$ is holomorphic in $B(0,2)\backslash\lbrace 0 \rbrace$. Expanding g in Laurent series, we have that
$g(z)=\sum_{n\in\mathbb{Z}}b_nz^n$ where $b_n:=\int_{\gamma}\frac{g(w)}{w^{n+1}}dw$
What I am trying to demonstrate is as follows: If $0$ is a pole for g then, for $n\geq 1$, $b_{-n}=a_n$.
This is what I have come to:
$\begin{align*} b_{-n} & = \frac{1}{2\pi i}\int_{\gamma}\frac{g(w)}{w^{-n+1}}dw \\\ & = \frac{-1}{2\pi i}\int_{\beta}\frac{g(w)}{w^{-n+1}}dw \\\ & = \frac{-1}{2\pi i}\int_{\beta}\frac{f(\frac{1}{w})}{w^{-n+1}}dw \\\ & = \frac{-1}{2\pi}\int_{0}^{2\pi}\frac{f(e^{it})}{(e^{it})^{n}}dt \\\ & = \frac{-1}{2\pi i}\int_{0}^{2\pi}\frac{f(e^{it})}{(e^{it})^{n+1}} i e^{it}dt \\\ & = \frac{-1}{2\pi i}\int_{0}^{2\pi}\frac{f(\gamma(t))}{(\gamma(t))^{n+1}} \gamma'(t)dt \\\ & = \frac{-1}{2\pi i}\int_{\gamma}\frac{f(w)}{w^{n+1}}dw \\\ & = - a_n \end{align*}$.
Is this correct?
At this point I have not been able to relate b_n with a_n, I would greatly appreciate your help.
P.D: I couldn't think of a better title, if someone has a better idea, feel free to modify it :)
You are almost there. Just substitute $t\gets-t$ in either of your final integrals.