Laurent series of square root of a polynomial Let $f(z) = \sqrt{z^2 -3z +2} = \sqrt{(z-1)(z-2)}, 1<|z|<2$. Does the Laurent series expansion for $f(z)$ exist? Justify your answer.
I'm quite new in Laurent series, so I do not know how to approach this. Thanks in advance!
 A: Gathering ideas from the discussion, I focus on a proof that $\sqrt{z^2-3z+2}$ does admit no Laurent expansion within the annulus
$$
\Omega:=\{z\in \mathbb{C}\ |\ 1<|z|<2\}
$$
firstly, remark that $\Omega$ being connex, $\mathcal{H}(\Omega)$ has no zrero divisors and any square root $f$ of $q(z)=z^2-3z+2$ being given, the set of solutions of $f^2=q$ is $\{-f,f\}$. Now,  following an idea of Oscar Lanzi, we define 
$$
g(z)=i\sqrt{1-1/z}\sqrt{2-z}=i\sqrt{2}\sqrt{1-1/z}\sqrt{1-z/2}
$$ 
and remark that, in the annulus $\Omega$, we have $|1/z|<1$ and $|z/2|<1$, then $g(z)$ admits a Laurent expansion using the expansion of $(1+Z)^{1/2}$ with, respectively, $Z=1/z$ and $Z=z/2$. Now $g(z)^2=1/z.q(z)$.
If $f(z)$ (any square root of $q$) admitted a Laurent expansion in $\Omega$, it would be the same for $h=f/g$. But, then $h^2(z)=z$ which is impossible as $z$ admits no square root within $\Omega$ (to prove this consider that $\rho e^{i\theta}\mapsto \rho e^{i\theta/2}$ is a solution of $u(z)^2=z$ defined on the open domain $\Re(z)>0$ which intersects $\Omega$).   
