I was studying Fourier Analysis: An Introduction by Stein and Shakarchi.
I came across this theorem in Section 2.2, page number 39
Theorem 2.1 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 $\theta_0$.
This was first proved for real-valued functions, which I understood. Then the proof was extended to complex-valued functions, with a brief explanation.
In general, write $f(\theta) = u(\theta) + iv(\theta)$, where $u$ and $v$ are real-valued. If we define $\overline f(\theta) = \overline{f(\theta)}$, then $$u(\theta)=\dfrac{f(\theta)+\overline f(\theta)}{2}$$ and $$v(\theta)=\dfrac{f(\theta)-\overline f(\theta)}{2i}$$ and since $\hat{\overline{f}}(n) = \overline{\hat f(-n)}$, we conclude that the Fourier coefficients of $u$ and $v$ all vanish, hence $f = 0$ at its points of continuity.
I have a confusion in the last part. How does $\hat{\overline{f}}(n) = \overline{\hat f(-n)}$ lead us to conclude that $f=0$ at its points of continuity?