Let $f(z)$ be a holomorphic function in an annulus $A(z_0,r_1,r_2)$ with $0\leq r_1\leq r_2$
Consider the Laurent expansion $f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$
How do I show that the analytic part and principal part of the series converges for $|z-z_0| <r_2$ and $|z-z_0|>r_1$ respectively?
What I tried to do:
For Analytic part; $\sum_{n=0}^{\infty} a_n (z-z_0)^n$
Consider $\Omega = D(0,r_2)$
$f(z)$ is holomorphic in $\Omega$ , so we can rewrite f(z) as
$f(z) = c_0 +c_1 (z-w) +c_2 (z-w)^2 + ... = \sum_{n=0}^{\infty} c_n (z-w)^n $ where the series converges for $|z-w|< r_2 $
Take $w=z_0$
Then $\sum_{n=0}^{\infty} c_n (z-w)^n = \sum_{n=0}^{\infty} c_n (z-z_0)^n =\sum_{n=0}^{\infty} a_n (z-z_0)^n $ by uniqueness of Laurent series
so we say that the Analytic part converges for $|z-z_0| <r_2$
For the principal part,
we do by the same argument?
Can somebody please tell me I'm on the right track? Or I should consider using $limsup$ etc to prove the convergence?