Power Series of a Holomorphic Function determined by its Real Part and $f(0)$? While looking at exercise sheets from last year, I encountered the following statement but wasn't able to prove it myself.
Let $f: D_R(0) \rightarrow \mathbb{C}$ be holomorphic and $ f(z)= \sum_{n=0}^{\infty} c_n z^n $ its power series.
Then for every $0 < r < R$ and $n \geq 1$: 
$$ c_n = \frac{1}{\pi r^n} \int_0^{2\pi} Re(f(re^{it}))e^{-int}dt $$
I know that we can write $c_n=\frac{1}{2\pi r^n} \int_0^{2\pi} f(re^{it})e^{-int}dt$.
Does this mean that a holomorphic function is uniquely determined by its real part and the value at zero? This looks like another "WTF?"-statement from complex-analysis to me...
 A: Since
$$\operatorname{Re} f(z) = \frac{1}{2}\bigl(f(z) + \overline{f(z)}\bigr),$$
we have
$$\frac{1}{\pi r^n}\int_0^{2\pi} \operatorname{Re} f(re^{it})e^{-int}\,dt = \frac{1}{2\pi r^n} \int_0^{2\pi} f(re^{it})e^{-int} + \overline{f(re^{it})}e^{-int}\,dt.$$
So it suffices to see that
$$\int_0^{2\pi} \overline{f(re^{it})}e^{-int}\,dt = 0$$
for $n \geqslant 1$. But
$$\overline{f(re^{it})}e^{-int} = \sum_{k=0}^\infty \overline{c_k} r^k e^{-i(k+n)t},$$
and we know
$$\int_0^{2\pi} e^{-i(k+n)t}\,dt = 0$$
for $k+n > 0$.

Does this mean that a holomorphic function is uniquely determined by its real part and the value at zero?

Indeed, on each domain, a holomorphic function is uniquely determined up to an imaginary constant by its real part, since a non-constant holomorphic function on a domain is an open mapping, thus holomorphic functions on a domain that attain only purely imaginary values are constant.
A: Once you know the real part $u(x, y)$ of a holomorphic function, you can solve for the imaginary part $v(x, y)$ up to a constant by Cauchy-Riemann equations. The value of the function at 0 determines that constant and hence the holomorphic function is completely determined.
