Coefficients for Laurent series expansion are unique Suppose $$f(z)=\sum_{n=-\infty}^\infty a_nz^n$$ for all $1&lt;|z|&lt;2$. Show that the coefficients $\{a_n\}$ are unique. 

I want to use the fact that the Laurent series of any function $f$ is unique. But to use that, I have to know that $f$ is holomorphic in $1&lt;|z|&lt;2$. So how can I do it then?
Suppose there's another expansion
$$f(z)=\sum_{-\infty}^\infty b_n z^n.$$
But then we would have
$$0=f(z)-f(z)=\sum_{-\infty}^\infty a_n z^n-\sum_{-\infty}^\infty b_n z^n=\sum_{-\infty}^\infty (a_n-b_n) z^n.$$
But for this to be true $\forall z$ in the domain, we must have $a_n-b_n=0, \forall n$. So, the coefficients, are unique.
Suppose there's another expansion
$$f(z)=\sum_{-\infty}^\infty b_n z^n.$$
But then we would have
$$0=f(z)-f(z)=\sum_{-\infty}^\infty a_n z^n-\sum_{-\infty}^\infty b_n z^n=\sum_{-\infty}^\infty (a_n-b_n) z^n.$$
But for this to be true $\forall z$ in the domain, we must have $a_n-b_n=0, \forall n$. So, the coefficients, are unique.