In Sheehan Olver's exposition of how Chebyshev series arise, he lets $f\in C^\infty[-1,1]$ and defines $$g(z)=f(J(z))$$ where $$J(z)=\frac{1}{2}\left(z+\frac{1}{z}\right)$$ is the Joukowsky map. He then claims that the Laurent series expansion of $g$, $$g(z)=\sum_{k=-\infty}^\infty \hat{g}_kz^k$$ satisfies $\hat{g}_k=\hat{g}_{-k}$. I tried proving this, but kept getting $\hat{g}_k=-\hat{g}_{-k}$: $$\hat{g}_k=\frac{1}{2\pi i}\oint\frac{g(z)}{z^{k+1}}\mathrm{d}z=\frac{1}{2\pi i}\oint g(1/z)(1/z)^{k+1}\mathrm{d}z=-\frac{1}{2\pi i}\oint g(\zeta)\zeta^{k-1}\mathrm{d}\zeta=-\hat{g}_{-k}$$ where the substitution $\zeta=1/z$ was used.
Where am I going wrong in my proof?
Edit: I initially linked to the wrong PDF. It is corrected now.
One small but important detail: orientation.
The substitution $\zeta = \frac{1}{z}$ reverses the orientation of the unit circle, for $z$ it was traversed counterclockwise, and for $\zeta$ it then is traversed clockwise. That gives you the sign.
Note that the uniqueness of the Laurent expansion gives the symmetry also without integration, since $g(z) = g(1/z)$, i.e.
$$\sum_{k=-\infty}^\infty a_k z^k = \sum_{k=-\infty}^\infty a_k (1/z)^k = \sum_{k=-\infty}^\infty a_k z^{-k} = \sum_{k=-\infty}^\infty a_{-k}z^k,$$
we must have $a_k = a_{-k}$ by the uniqueness of the Laurent expansion.