Trigonometric Polynomial Coefficients Suppose $a(z)=\sum_{j=-n}^n a_j z^j \geq 0$ on the unit circle $|z|=1$. I would like to prove the seemingly simple fact that $a_j=\overline{ a_{-j}}$.
My attempt:
\begin{align}
a(e^{i\theta}) = \sum_{j=-n}^n a_j e^{ij\theta} \geq 0 \\
a(e^{-i\theta}) = \sum_{j=-n}^n a_j e^{-ij\theta} \geq 0 \\
\implies a(e^{i\theta}) \pm a(e^{-i\theta}) \in \mathbb R
\end{align}
Since $e^{i\theta}+e^{-i\theta}=2\cos(\theta)$ and $e^{i\theta}-e^{-i\theta}=2i\sin(\theta)$ we conclude that $\forall\theta$,
\begin{align} \tag{1}
\sum_{j=1}^n(a_j+a_{-j})\cos(j\theta) \in \mathbb R\\
\sum_{j=1}^n(a_j-a_{-j})\sin(j\theta) \in i\mathbb R
\end{align}
Of course,
\begin{align}\tag{2}
\begin{cases}
\alpha+\beta\in\mathbb R\\
\alpha-\beta\in i\mathbb R
\end{cases}
\iff
\alpha=\overline\beta
\end{align}
But I'm just not managing to move from the sums in $(1)$ to each equality in the form of $(2)$. The key is in the fact that $(1)$ holds for all $\theta$, but I'm still missing that crucial step that seems like "equating coefficients". Thanks!
 A: Orthogonality.
$\int_0^{2\pi} \sin(nx)\sin(mx)\ dx = \pi\cdot \delta_{n,m}$
A direct proof can be done simply,
\begin{align}
\sin(x) &= \frac {e^{ix}-e^{-ix}} {2i} \\
\cos(x) &= \frac {e^{ix}+e^{-ix}} {2} \\
\implies \sin(nx)\sin(mx) &= \frac 1 {-4}(e^{inx}-e^{-inx})(e^{imx}-e^{-imx}) \\
&= \frac 1 {-4}(e^{i(n+m)x}-e^{i(n-m)x} -e^{-i(n-m)x}+e^{-i(n+m)x}) \\
&= \frac 1 {-4}(2\cos((n+m)x)-2\cos((n-m)x)) \\
\implies \int_0^{2\pi} \sin(nx)\sin(mx)\ dx &= \frac 1 2 \int_0^{2\pi} \cos((n-m)x) - \cos((n+m)x)\ dx \\
\therefore n=m \implies \int_0^{2\pi} \sin(nx)\sin(mx)\ dx &= \frac 1 2 \int_0^{2\pi} 1 - \cos(2nx)\ dx = \pi\\
n\neq m \implies \int_0^{2\pi} \sin(nx)\sin(mx)\ dx &= \frac 1 2 \int_0^{2\pi} \cos(\tilde m x) - \cos(\tilde nx)\ dx = 0\\
\end{align}
Therefore we reduce our sum by the standard $L_2([0,2\pi])$ inner product (technically we have to split real and imaginary and apply separately):
\begin{align}
\sum_{j=1}^n(a_j+a_{-j})\cos(j\theta) &\in \mathbb R
\implies \frac 1 \pi \langle \sum_{j=1}^n(a_j+a_{-j})\cos(j\theta), \cos(k\theta) \rangle =  a_k+a_{-k} \in \mathbb R \\
\sum_{j=1}^n(a_j-a_{-j})\sin(j\theta) &\in i\mathbb R
\implies \frac 1 \pi \langle \sum_{j=1}^n(a_j-a_{-j})\sin(j\theta), \sin(k\theta) \rangle = a_k-a_{-k} \in i\mathbb R \\
\end{align}
