An analytic function satisfies $f(1/z)=f(z) $, if $f$ is real on $\{|z|=1\}$, then the coefficients of expansion are real. An analytic function satisfies $f(1/z)=f(z),\forall z \in \mathbb{C}\backslash\{0\} $, if $f$ has real values on $\{|z|=1\}$, then the coefficients of the Laurent expansion are all real and.
Here is my try, $f(z)=\sum_{n=-\infty}^{n=\infty} \alpha_n z^n=f(\frac{1}{z})=\sum_{n=-\infty}^{n=\infty} \alpha_nz^{-n}$, so $\alpha_n=\alpha_{-n},\forall n\geq 1.$
On $|z|=1, f(z)$ is real, so $\alpha_0=\frac{1}{2\pi i}\int_{|\zeta|=1}\frac{f(\zeta)}{\zeta}d\zeta=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}} \cdot e^{i\theta}d\theta\in \mathbb{R}$.
When $n\geq 1$,
$\alpha_n=\frac{1}{2\pi i}\int_{|\zeta|=1}\frac{f(\zeta)}{\zeta^{n+1}}d\zeta=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta {(n+1)}}} \cdot e^{i\theta}d\theta=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta n}}d\theta$
$\bar{\alpha_n}=\overline{\frac{1}{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta n}}d\theta}=\frac{1}{2\pi}\int_0^{2\pi}\overline{(\frac{f(e^{i\theta})}{e^{i\theta n}})}d\theta=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{-i\theta n}}d\theta=\alpha_{-n-1}$
Actually the answer said that $\bar{\alpha_n}=\alpha_{-n}$. I don't know what is wrong in the calculation.
 A: The expression you compute for $\overline{a_n}$ looks much more like the expression you computed for $a_{-n}$ thanlike the one you computed for $a_{-n-1}$.
A: An approach which is quite similar, but not completely identical, to that of Adam Hughes:  consider $z$ such that $\vert z \vert = 1$; we may write $z = e^{i \theta}$ for such $z$.  Then we have $\bar z = z^{-1} = e^{-i \theta}$.  Expanding $f(z) = f(z^{-1})$ in terms of $e^{i \theta}$ on the circle $\vert z \vert = 1$ yields
$f(z) = f(e^{i \theta}) = \sum_{n = -\infty}^{n = \infty} a_n e^{in\theta} \tag{1}$
and
$f(z^{-1}) = f(e^{-i \theta}) = \sum_{n = -\infty}^{n = \infty} a_n e^{-in\theta}; \tag{2}$
using $f(z) = f(z^{-1})$ we obtain
$\sum_{n = -\infty}^{n = \infty} a_n e^{in\theta} = \sum_{n = -\infty}^{n = \infty} a_n e^{-in\theta}, \tag{3}$
from which it follows that $a_m = a_{-m}$ for all $m \in \Bbb Z$, as has been observed by our OP user159895.  Or, to make a somewhat more formal argument, we note that, since $f(z)$ is analytic on $\Bbb C \setminus \{0\}$, it is continuous on $T = \{z \in \Bbb C \mid \vert z \vert = 1 \}$, hence $f(e^{i\theta}) \in L^2(T)$, and since the functions $e^{in\theta}$, $-\infty < n < \infty$, constitute an orthonormal basis for $L^2(T)$ with respect to the usual inner product
$\langle \omega, \sigma \rangle = \dfrac{1}{2\pi}\int_0^{2\pi} \bar \omega(\theta) \sigma(\theta) d\theta, \tag{4}$
we may simply compute, from (3)
$a_m = \langle e^{im\theta}, \sum_{n = -\infty}^{n = \infty} a_n e^{in\theta}\rangle = \langle e^{im\theta}, \sum_{n = -\infty}^{n = \infty} a_n e^{-in\theta} \rangle = a_{-m}; \tag{5}$
likewise, exploiting the hypothesis that $f(z)$ is real on $T$, so that $f(z) = \overline {f(z)}$ there, we find that
$\sum_{n = -\infty}^{n = \infty} a_n e^{in\theta} = \overline{\sum_{n = -\infty}^{n = \infty} a_n e^{in\theta}} = \sum_{n = -\infty}^{n = \infty} \bar a_n e^{-in\theta}; \tag{6}$
if we now take the $L^2$ inner product of (6) with $e^{-im\theta}$ we see that
$\bar a_m = a_{-m}, \tag{7}$
which combined with our previous result
$a_m = a_{-m} \tag{8}$
yields
$\bar a_m = a_m, \tag{9}$
the desired conclusion.  QED
Note:  The preceding answer assumes my readers are familiar with the basic facts of Fourier analysis on the unit circle $T$, such as $\langle e^{im\theta}, e^{in\theta} \rangle = \delta_{mn}$ and $c_m = \langle e^{im\theta}, \sum_{n = -\infty}^{n = \infty} c_n e^{in\theta} \rangle$ for $\sum_{n = -\infty}^{n = \infty} c_n e^{in\theta} \in L^2(T)$.  The former is easy to see by direct integration:
$\langle e^{im\theta}, e^{in\theta} \rangle = (1/2\pi)\int_0^{2\pi} e^{-im\theta}e^{in\theta}d\theta = (1/2\pi)\int_0^{2\pi} e^{i(n - m)\theta}d\theta$
$= (1/2\pi)(i(n - m))^{-1}(e^{i(n - m)2\pi} - e^{i(n - m)(0)}) = 0 \tag{10}$
if $m \ne n$ and
$\langle e^{n\theta}, e^{in\theta} \rangle = (1/2\pi)\int_0^{2\pi} e^{i(n - n)\theta}d\theta = (1/2\pi)\int_0^{2\pi}1 d\theta = (2\pi/2\pi) = 1. \tag{11}$
The formula
$c_m = \langle e^{im\theta}, \sum_{n = -\infty}^{n = \infty} c_n e^{in\theta} \rangle \tag{12}$
follows from (10) and (11) via term-by-term integration.  But these are commonly known tricks of the trade, and most likely anyone reading this post has seen them already.  End of Note.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
