Given that a conformal map on the disk $\mathbb{D}$ will always have the form $f(z)=\lambda \displaystyle\frac{z-w}{1-\overline{w}z}$ for some $\lambda\in \partial \mathbb{D}$ and some $w\in \mathbb{D}$, what are the Fourier coefficients for $f(z)$ where the Fourier coefficients are given by $c_k=\displaystyle \int_{-\pi}^\pi f(e^{i\theta})e^{-ik\theta}\frac{d\theta}{2\pi}=\lambda\int_{-\pi}^\pi \frac{e^{i\theta}-w}{1-\overline{w}e^{i\theta}}e^{-ik\theta}\frac{d\theta}{2\pi}$?
Since $f(z)$ is analytic, $c_k$ should be $0$ for $k<0$. Since and inner map on the disk corresponds to an isometric Toeplitz operator, I highly suspect that $c_k=0$ for all $k\neq n$ for some $n$ (and $|c_n|=1$). But, I can't figure out any nice way of writing the integral to say this.
The best I can say is something like $c_k= \displaystyle \lambda\int_{-\pi}^\pi \frac{e^{i\theta}-w}{1-\overline{w}e^{i\theta}}e^{-ik\theta}\frac{d\theta}{2\pi}=\lambda\int_{-\pi}^\pi \frac{e^{i\theta}-w}{e^{-i\theta}-\overline{w}}\frac{e^{-ik\theta}}{e^{i\theta}}\frac{d\theta}{2\pi}=\lambda\int_{-\pi}^\pi \frac{e^{i\theta}-w}{\overline{e^{i\theta}-w}}e^{-i(k+1)\theta}\frac{d\theta}{2\pi}$,
which doesn't seem to get me anywhere.
Any ideas on how to integrate this?
In fact, you are better off sticking with $z$, or more specifically, from $$ \frac{1}{2\pi} \int_{-\pi}^{\pi} f(e^{i\theta}) e^{-ik\theta} \, d\theta, $$ set $z=e^{i\theta}$, so $ dz/(iz) = d\theta $, and the integral becomes $$ \frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z^{k+1}} \, dz. $$ Now, $f$ is analytic, so the Residue theorem (or the Cauchy integral formula) shows that this integral is equal to $ \frac{1}{k!}f^{(k)}(0) $ for $k \geqslant 0$ and zero for $k<0$, because when $k<0$ the whole integrand is analytic and Cauchy's theorem says the integral is zero (for $k \geqslant 0$, use the Taylor expansion, and so on).
So actually, you can just series expand $\lambda \frac{z-w}{1-\bar{w}z}$ and find the coefficient of $x^k$ to find the Fourier series. This works for any function analytic on the disc (from this you can, through careful definitions, find the Fourier series of the sawtooth function and the square wave, for example).
The functional analysis properties you refer to are presumably manifest in the negative coefficients being zero.