On page 40, the book tries to prove uniqueness of fourier series the following: Suppose that $f$ is an integrable function on the circle with $\hat{f}(n)=0$ for all $n \in \mathbb{Z}$. Then $f(\theta_{0})=0$ whenever f is continuous at the point $\theta_{0}$.
To prove this, assuming that $f$ satisfies the hypotheses, $\theta_{0}=0$, and $f(0)>0$. The idea is now to construct a family of trigonometric polynomials $\{p_{k}\}$ that peak at $0$, resulting in $$\int p_{k}(\theta)f(\theta)\; d\theta \rightarrow \infty \;\; \text{as}\;\; k \rightarrow \infty.$$ And this will contradict the requirement that the coefficient needs to be 0.
To construct the p, since $f$ is continuous at $0$, we can choose $0 < \delta \leq \pi/2$, so that $f(\theta)>f(0)/2$ whenever $|\theta|< \delta$. Let $p(\theta) = \epsilon + \cos\theta$ where $\epsilon >0$ is chosen so small that $|p(\theta)|< 1-\epsilon/2$, whenever $\delta \leq |\theta| \leq \pi$. Then choose a positive $\eta$ with $\eta < \delta$, so that $p(\theta) \geq 1 + \epsilon/2$, for $|\theta| < \eta$. Finally, we define $p_{k}(\theta) = [p(\theta)]^k$.
Then the key point will be when $|\theta| < \eta$,$$\int p_{k}(\theta)f(\theta)\; d\theta \geq 2\eta \frac{f(0)}{2}(1+\frac{\epsilon}{2})^k \rightarrow \infty \; , k \rightarrow \infty$$ This contracts that the coefficient needs to be 0. So the proof finishes here.
What I'm not understanding is when computing the fourier coefficient using normal formula, $$\hat{f}(n)=\int f(\theta)e^{-in\theta}\; d\theta.$$ If I sum over multiple coefficients, the formula will be
$\int f(\theta)(e^{-i\theta}+e^{-i2\theta}+e^{-i3\theta}+...)\; d\theta.$
it will not be in the form of $p_k(\theta) = (\cos\theta + \epsilon)^k $. Unless $p_k(\theta) = (\cos\theta + \epsilon)^k = e^{-i\theta}+e^{-i2\theta}+e^{-i3\theta}+... $. under some conditions?