Exist complex $z_{0}$ ,such $|z_{0}|=1$,and $|f(z_{0})|\le|f(z)|,\forall |z|\ge 1$ Let $a\in (0,1), f(z)=z^2-z+a, z\in \mathbb C$. Does there exist a complex number $z_{0}$ such that $|z_{0}|=1$, and 
$$|f(z_{0})|\le|f(z)|,\forall |z|\ge 1$$
I just have no idea where to even begin so any hint will be much appreciated. I apologize for not showing any effort. Any help will be appreciated. Thanks
Is there a method without using the Maximum Modulus Principle to solve this problem?
 A: We have $\lim\limits_{z\to\infty} |f(z)|=\infty$ and the function $|f|$ is continous, so $|f|$ has a minimum at some point $z_0$ in the set $|z|\ge1$. Our goal is to prove that $z_0$ is located on the unit circle; in other words, $z_0$ cannot be outside the circle.
Let the two roots be $u$ and $v$. They are in the closed unit disk: $u,v=\frac12\pm\sqrt{\frac14-a}\in[0,1]$ if $0\le a\le\frac14$, and 
$u,v=\frac12\pm\sqrt{a-\frac14}i$ if $\frac14\le a\le1$. We know that
$$ f(z) = |z-u| \cdot |z-v|. $$
From this point it is just some basic geometry. Suppose that $z_0$ is outside the circle and consider the (may be degenerate) triangle formed by $z_0$, $u$ and $v$.
Take a point $w$, close to $z_0$, on the angle bisector starting at $z_0$, but outside the unit circle. The angles $\angle u,z_0,w$ and $\angle v,z_0,w$ are acute, so $|w-u|<|z_0-u|$ and $|w-v|<|z_0-v|$; therefore $|f(w)|<|f(z_0)|$; so $|f|$ cannot have a minimum at $z_0$.
A: The function $|f|$ is continuous and $\lim_{|z|\to\infty } |f(z)|=\infty $ hence there exists $z_0 $ such that $$\inf_{z\mathbb{C} \setminus U}|f(z)| =|f(z_0 )|$$
where $U=\{z:|z|<1\}.$
