Showing $ \int_0^{2\pi} f(e^{it}) e^{it} dt =0$ for holomoprhic function $f$

Question

Let $U\subseteq \mathbb{C}$ be an open set. Suppose that $U\supseteq \{ e^{it} : 0 \leq t <2\pi \}$. Let $f$ be a holomorphic function on $U$ which has a holomorphic antiderivative. then we have \begin{align} \int_0^{2\pi} f(e^{it}) e^{it} dt =0 \end{align}

I have no clue for this, can you give me some proof or brief idea for this problem?

Answer 1

You could consider: $$i\int_{0}^{2\pi}f(e^{it})e^{it}dt$$

Which is: $$\int_{C} f(z)dz$$

If f is holomorphic in $C$ (assuming that $C(t) = e^{it}, 0\le t \le2\pi$) then that integral is equal to $0$ by the residue theorem. Then: $$\int_{0}^{2\pi}f(e^{it})e^{it}dt = 0$$

because it's the same integral without the $i$ infront of it.