Let,
$f\in H(B(0,R))$ and $f\in C(B[0,R])$, let $$ T_R(t)=Re^{it}$$,
My problem is to show that$$\int_{T_R}f=0$$
It is easy to say, for any closed path inside the circle, the value of integration is $0$, by Cauchy theorem. i.e $$\int_{T_r}f=0, r<R$$.
But How to say the integration value is zero even on the boundary?
Hint: the integral over the circle of radius $R-\epsilon$ is $0$ for every $\epsilon >0$. Use uniform continuity of $f$ on the closed disk to prove that this integral tends to the integral over the circle of radius $R$ as $\epsilon \to 0$.