Every Holomorphic function on a Hartog's figure can be extended holomorphically to the while of $P^n$ Let $f$ be a holomorphic function on the Euclidean Hartog's figure that is $$H=\{(z,w)\in P^2 : 1 \gt|z| \gt q_1 \text{or} |w|  \lt q_2 \}$$ where $0 \lt q_i \lt 1$. I need to show that it has a holomorphic extension to whole of $P^2$ where $P^2$ is the unit polydisc. 
For a fixed $z$, $$f(z,w)=\sum_{n=-\infty}^{\infty}a_n(z)w^{n}$$, where $$a_n(z)=\frac{1}{2\pi i}\int_{|\alpha|=r} \frac{f(z,\alpha)}{\alpha^{n+1}}d\alpha$$ where $0 \lt r \lt q_2$. Since $f$ depends continuously on $z$ (How do I prove it??) and $f(.,\alpha)$ is holomorphic in $|\alpha| \lt q_2$ we get $$\frac{\partial a_n(z)}{\partial \bar{z}}=\frac{1}{2\pi i}\int_{|\alpha|=r} \frac{\partial f(z,\alpha)}{\partial \bar{z}}.\frac{1}{\alpha^{n+1}}d\alpha=0$$ This shows that $a_n(z)$ is a holomorphic function of $z$ for $|z| \lt 1$. 
Now for $|z| \gt q_1$, $a_n(z)=0$ for $n \le -1$ since $f$ is holomorphic in this region. Now since $a_n(z)$ is holomorphic , $a_n(z)=0$ for $n \lt -1$ in the whole $|z| \lt 1$.
 And we are done. 
Is this alright??
Thanks for the help!!

You don't need the Laurent series expansion of $f$. Just extend your function via the Cauchy integral formula in the variable $z$ and show that the extension is complex-differentiable along each coordinate direction.
