Need help with problems in complex analysis about analyticity Show that $\nexists f\in H(B_1(0))\cap C(\bar{B}_1(0))$ such that $f(z)=\bar{z}$ on $\delta B_1(0)$ 
i.e., there exists no $f(z)$ analytic on open unit circle and continuous on closure of unit circle such that $f(z)=\bar{z}$ on the boundary of unit circle. 
I think I have to use schwarz reflection principle here, but don't know how?
 A: By Cauchy's formula, one has
$$ f^{(n)}(0) = \frac {n!}{2i\pi} \int_{rS^1} \frac {f(z)}{z^{n+1}} dz,$$
for any $r < 1$.
Since $f$ is continuous on the closed disk, you can let $r$ tend to $1$ in the equality. The corresponding integral 
$$ \int_{S^1} \frac 1{z^{n+2}} dz $$ 
is $0$ for all natural $n$. Hence, $f$ is the zero function, which is not possible.
Edit: we have $$f^{(n)}(0) = \frac {n!}{2i\pi} \int_{S^1} \frac {f(z)}{z^{n+1}} dz = \frac {n!}{2i\pi} \int_{S^1} \frac {1}{z^{n+2}} dz.$$
(since $\bar{z} = \frac 1z$ on the circle)
The last integral is equal to
$$\int_0^{2\pi} ie^{i\theta} \times e^{-i(n+2)\theta} d\theta$$
and this is zero.
A: Basically the same idea, but a little quicker:
$$
\int_{|z|=1} \bar z\,dz = \int_{|z|=1} \frac{z\bar z}{z} \,dz = \int_{|z|=1} \frac{1}{z} \,dz = 2\pi i,
$$
either by direct computation or by referring to Cauchy's integral formula. On the other hand, if it was possible to extend $\bar z$ holomorphically, then the integral would be $0$ by Cauchy's integral theorem. Hence, no such extension can exist.
