Analytic continuation from punctured disk to open disk

Question

Let $f(z)$ be a complex-valued function that is analytic on the punctured disk $0 < |z-w| < d$ and suppose that $f$ is continuous at $w$. Is $f$ analytic at $w$ itself?

Answer 1

Hint: Define $g(z)=(z-w) f(z)$. Show that $g$ is analytic on $0< |z-w|

Therefore, $g(z)$ has a Taylor series.

Use $$g(w)=\lim_{z \to w} g(z)$$ to deduce that $g(w)=0$. use this to show that $f$ has a Taylor series at $w$.

Alternate Approach By Laurent theorem, $$f(z)=\sum_{n=0}^\infty a_n (z-w)^n + \sum_{n=1}^\infty b_n \frac{1}{(z-w)^n}$$ where $$b_n =\frac{1}{2 \pi i} \int_C f(z) (z-w)^{n-1} $$

Show as above that $f(z) (z-w)^n$ is differentiable at $z=w$ and therefore Analytic on the entire disk. What does this say about $b_n$?

Once you prove that $b_n=0$ for $n \geq 2$, deduce that $$f(z)= b_1 \frac{1}{(z-w)}+\sum_{n=0}^\infty a_n (z-w)^n \\ (z-w)f(z)= b_1+\sum_{n=0}^\infty a_n (z-w)^{n+1} $$

Now take the limit when $z \to w$ to deduce that $b_1=0$.