Let, $f$ be analytic in $D\setminus \{a\}$. Also, $f$ is continuous at $z=a$. Then prove that $f$ is analytic on whole of $D$.
We know from Riemann's Theorem, if $f$ is not analytic at $a$ and in a nbd. of $a$, if $f$ is bounded then either $f$ is analytic at $a$ or $f$ has removable singularity at $a$. But here how can I use the continuity condition to show $f$ is analytic ?
It is not restrictive to assume that $a=0$ (with a translation).
Consider $g(z)=z^2f(z)$, for $z\in D$, $z\ne0$, and $g(0)=0$.
Prove that $g$ is differentiable at $0$. Thus it is differentiable over $D$, hence analytic over $D$. Hence it has a Taylor series expansion at $0$: $$ g(z)=\sum_{n=0}^{\infty}b_nz^n $$ but $b_0=b_1=0$ (prove it). Hence $$ f(z)=\sum_{n=0}^{\infty}b_{n+2}z^n $$ is analytic in a neighborhood of $0$.
(Adapted from Wikipedia)