Suppose $p(z)$ and $q(z)$ are complex polynomials such that all of $p$'s zeros are in the open unit desk and they have the same degree. Prove that if $|q(z)| \leq |p(z)|$ for $|z|=1$, then $|q(z)| \leq |p(z)|$ for $|z| > 1$.
My attempt: I considered $g = \frac{q}{p} (z)$ being holomorphic in $|z| > 1$. Then $g$ attains a maximum the boundary, i.e. for $|z|=1$ or $ { \infty }$. We have that $|g(z)| \leq 1$ on $|z|=1$ - and since $p$ and $q$ have the same degree - $|g(z)| = a$ on |z| approaching $\infty$ (where $a$ is the magnitude of the quotient of the leading coefficients of $q$ and $p$). Hence, to show that $|g(z)| \leq 1$ for $|z| > 1$ we must show that $a \leq 1$.
Here, I'm a little stuck. I can show $ a \leq 1$ when the degree of $p$ and $q$ is 1, but this argument doesn't generalize well.
Context: Practicing for my Quals ( Number 6 in http://www.math.tamu.edu/graduate/phd/quals/ncomplex/a13.pdf )
You were on the right path. $g = q/p$ is holomorphic in $1 \le |z| < \infty$, with a removable singularity at $z = \infty$, and therefore is bounded by its values on $|z| = 1$.
If you feel uncomfortable with a holomorphic continuation at $z=\infty$ then consider $$ h(z) = \frac{q(1/z)}{p(1/z)} $$ instead, which has a removable singularity at $z=0$.
(Note that it suffices to require $\deg q \le \deg p$ instead of $\deg q = \deg p$.)