Cauchy goursat's theorem and holomorphicity If f is continuous on the unit disc and holomorphic on the punctured disc, why does this imply that f is holomorphic on the entire disc?
 A: Write a Laurent series for $f$ in the annulus $0<|z|<1$. If $f$ is continuous at $0$, what does that tell you about the coefficients of $z^{-n}$, $n=1,2,\dots$?
A: As long as $f$ is bounded in a punctured neighbourhood of $0$, the singularity at $0$ is removable.
A: In complex analysis we can't do anything without the Cauchy integral theorem. Here is how you can use it :
If $g(z)$ is holomorphic on $U = \{z \in \mathbb{C}, \ 0< |z| < 1, z \ne s\}$ then for $r < 1-\epsilon, \epsilon < |s| < r-\epsilon$ the contour $\{|z| = r\}$ is homotopically equivalent to $\{|z| = \epsilon\} \ \ \cup \ \ \{|z-s| =\epsilon\}$ on $U$, so that
$$\int_{|z| = r} g(z)dz=\int_{|z| = \epsilon} g(z)dz+\int_{|z-s| = \epsilon} g(z)dz$$
Applying it to $\displaystyle\frac{f(z)}{z-s}$, $0 < |s| < r$  you get
$$\int_{|z| = r}\frac{f(z)}{z-s}dz = \lim_{\epsilon\to 0} \int_{|z| = \epsilon}\frac{f(z)}{z-s}dz+\int_{|z-s| = \epsilon}\frac{f(z)}{z-s}dz\qquad\qquad $$ $$\qquad\qquad=\lim_{\epsilon\to 0}\int_{|z| = \epsilon}(\frac{f(0)}{-s}+o(1))dz +\int_{|z-s| = \epsilon}\frac{f(s)+o(1)}{z-s}dz = 2i \pi f(s)$$
The LHS is clearly holomorphic (and analytic) in $s$ on $|s| < r$, hence so is the RHS.
