Proving $f(z)$ entire function in complex analysis If $f\in C(\Bbb C)\cap H(\Bbb C\backslash \delta B_1(0))$ then $f\in H(\Bbb C)$ [C means continuous, $\Bbb C$ means complex plane, H means analytic and $\delta$ means boundary] 
I don't even know where to start. Thank you...
 A: Almost certainly you can prove this with Morera's theorem. 
In particular, I think this sketch works. Suppose you have some closed simple curve that cuts $\partial\mathbb D$. (Actually Morera's theorem is true if you just test on rectangles, which circumvents the problem of dealing with nasty curves.) Then adding in the bits of the circle the curve cuts out, you get two closed curves, one lying in $\overline{\mathbb D}$ and one lying in $\overline {\mathbb D}^c$, that sum to your original curve. It suffices to show that the integral over each of these two curves is zero.
For the curve $\gamma$ inside the circle, consider the corresponding curve $\gamma_t$ you get by scaling $\gamma$ by $t\in(0,1)$ around zero. Each $\gamma_t$ is zero, and the function $t\rightarrow \int_{\gamma_t}f \ dz$ should be continuous, so the integral of $f$ over $\gamma$ is $0$. A similar argument should work for the other curve. 
A: Since functions which are holomorphic everywhere except on a line are easier to deal with, let us avoid dealing with the boundary of the circle directly.
Consider the map $g=f\circ c$, where $c$ is any Mobius transformation on $\mathbb{C}\cup \lbrace \infty \rbrace$ sending $\delta B_1(0)$ to $\mathbb{R}$; e.g., $c(z)=i\frac{1+z}{1-z}$. Then $g$ is holomorphic on $\mathbb{C}-\mathbb{R}$ and continuous on $\mathbb{C}$. Apply the argument using triangles on $\mathbb{C}-\mathbb{R}$ given by the answer in Function holomorphic except for real line and continuous everywhere is entire to conclude that $g$ is entire. Since $c$ is an entire bijection on $\mathbb{C}$, so is $c^{-1}$ and thus so is the composition $g\circ c^{-1}=f$, as desired.
