Theorem 2.1 in Stein and Shakarachi the complex valued function case. The book said "In general, write $ f(\theta)= u(\theta) + iv(\theta) $, where u and v are real valued. if we define $\bar{f}(\theta) = \overline{f(\theta)},$ (really I do not know what is the meaning and importance of this sentence)then
$u(\theta) =\frac{f(\theta) + \bar{f}(\theta)}{2}$ and $v(\theta) =\frac{f(\theta) - \bar{f}(\theta)}{2i}$, and since $\hat{\bar{f}}(n) =\overline{ \hat{f}(-n)}$, we conclude that the Fourier Coefficients of u and v all vanish."
I have made some calculations with the information given but this does not lead to that the Fourier Coefficients of u and v all vanish. Could anyone help me please?
thanks. 

For reference, the "theorem 2.1" in question is

Theorem: Suppose that $f$ is an integrable function on the circle with $\hat{f}(n) = 0$ for all $n \in \Bbb{Z}$. Then $f(\theta_0) = 0$ whenever $f$ is continuous at the point $\theta_0$.

The authors first prove the result for real-valued $f$.
Now let $f$ be an integrable complex-valued function on the circle with $\hat{f}(n) = 0$ for all $n \in \Bbb{Z}$. Suppose that $f$ is continuous at the point $\theta_0$ and that
$$
f(\theta) = u(\theta) + iv(\theta)
$$
where $u$ and $v$ are real-valued functions.
Then $u$ and $v$ are also integrable functions on the circle and continuous at the point $\theta_0$.
In order to apply the real-valued case of the result, we need to make sure that the Fourier coefficients of $u$ and $v$ all vanish.
Recall that for any complex number in cartesian form $z = a+ib$, we have
$$
a = \frac{z+\overline{z}}{2} \\
b = \frac{z-\overline{z}}{2i}
$$
where $\overline{z} := a-ib$ is the complex conjugate of $z$. Hence
$$
u(\theta) = \frac{f(\theta)+\overline{f(\theta)}}{2} \\
v(\theta) = \frac{f(\theta)-\overline{f(\theta)}}{2i}
$$
The function $\overline{f}(\theta) := \overline{f(\theta)}$ is also an integrable function on the circle so its Fourier coefficients are defined. Note that
\begin{align}
\hat{\overline{f}}(n) &= \frac{1}{2\pi} \int_0^{2\pi} \overline{f}(t) e^{-int} \,dt\\
&= \overline{\frac{1}{2\pi} \int_0^{2\pi} f(t) e^{-i(-n)t} \,dt}\\
&=\overline{\hat{f}(-n)}
\end{align}
Thus
$$
\hat{u}(n) = \frac{\hat{f}(n)+\overline{\hat{f}(-n)}}{2} \\
\hat{v}(n) = \frac{\hat{f}(n)-\overline{\hat{f}(-n)}}{2i}
$$
By hypothesis, $\hat{f}(n) = 0$ for all $n$, hence $\hat{u}(n) = 0$ and $\hat{v}(n) = 0$ for all $n$.
By the real-valued case of the result, it follows that $u(\theta_0) = 0$ and $v(\theta_0) = 0$. Finally,
$$
f(\theta_0) = u(\theta_0) + iv(\theta_0) = 0
$$
