Given complex polynomial with roots within an annulus, there exists its "square root" which is analytic outside this annulus 
We are given a polynomial $f(z)$ of degree $2n$, which has all its roots within the annulus $|z|<a$. Now we must show that there is an analytic function $g(z)$ which is defined on $|z|>a$ which has the property that $g(z)^2=f(z)$.

I have no idea what property I could use to show this is true (I don't even know why it is true). I think there could be some significance to the degree of the polynomial being even though - if it were odd, say degree $1$, then the square root can't possibly be a polynomial. I don't know how to formalise this, and more importantly I don't see the significance of the regions. 
Could someone please explain this to me?
 A: Suppose the roots of $f$ are in $|z| < a$, and define
$$
            g(z) = C\exp\left\{\frac{1}{2}\int_{a}^{z}\frac{f'(w)}{f(w)}dw\right\},
$$
where $C^2=f(a)$, and where the path from $a$ to $z$ lies in $|z| \ge a$. The definition does not depend on the particular choice of the contour because, if $z_1,\cdots,z_k$ are distinct roots of $f$ with multiplicities $r_k$, then $\sum_k r_k = 2n$, and it follows that
$$
              \oint_{|z|=a}\frac{f'(w)}{f(w)}dw = \oint_{|z|=a}r_k\frac{1}{w-z_k}dw=2\pi i\sum_k r_k = 4\pi i n.
$$
So the particular choice of contour from $a$ to $z$ does not affect the value of $g(z)$. The function $g$ extends and is holomorphic in $|z| > a-\epsilon$ for some small $\epsilon$; and $g(a)^2=f(a)$ holds by design. Both $f,g$ are non-vanishing in $|z| > a-\epsilon$. And one may verify that
$$
     \frac{g'(z)}{g(z)}=\frac{1}{2}\frac{f'(z)}{f(z)},\;\; |z| > a-\epsilon
$$
Therefore,
$$
    \frac{d}{dz}\frac{f}{g^2} = \frac{g^2f'-2gg'f}{g^4}=\frac{2f}{g^2}\left[\frac{1}{2}\frac{f'}{f}-\frac{g'}{g}\right] = 0,\;\; |z| > a-\epsilon
$$
So $f=Cg^2$ in $|z| > a-\epsilon$ for some constant $C$. The constant $C$ is $1$ because $f(a)=g(a)^2$.
A: Write
$$f(z)=b_0 z^{2n}+b_{1}z^{2n-1}+\cdots+b_{2n}
=c^2z^{2n}F(1/z)$$
where $c$ is a square root of $b_0$ and
$$F(z)=1+\frac{b_1}{b_0}z+\cdots+\frac{b_{2n}}{b_0}z^{2n}
=\prod_{k=1}^{2n}(1-\alpha_k z).$$
Now $F(z)$ will be nonzero for $|z|<1/a$ so each $|\alpha_k|<a$.
Each $(1-\alpha_kz)$ will have a holomorphic square root on the disc
$\{z:|z|<1/a\}$.
