Given f is analytic in D and satisfy |f(z)|->1 as |z|->1 then prove that f is rational Let $f$ be analytic in $D$ and satisfy $|f(z)|\rightarrow 1$ as $|z|\rightarrow 1$. Prove $f$ is rational. I don't know how to go about this. I don't need Blachke products to verify this. Is there a simpler way? I do know that if we have $\phi$ is one-to-one analytic map of $D$ onto $D$, where $\phi (z)=\frac{z-a}{1-wz}$, then would that serve as a rational function that satisfies the conditions?
I am not sure how to proceed. Anyone know how to solve it?
 A: The standard proof of this involves noticing that the hypothesis implies that $f$ has only finitely many zeroes in the unit disc (as $|f| > 1/2$ for $|z|$ close enough to $1$), so one cn take the corresponding Blaschke product $B$ with the same zeroes and notice that $f/B, B/f$ do not vanish and $|f/B| \to 1, |B/f| \to 1, |z| \to 1$ hence by maximum modulus $|f/B| \le 1, |B/f| \le 1$ in the unit disc, hence $|f/B|=1$ there, $f=\alpha B, |\alpha|=1$
Edit later - as per comments, I thought of a different way of doing it as I realized that the hypothesis implies that $u=\log |f|$ is harmonic in a neighborhood $r <|z| <1$ (as before $f$ has only finitely many zeroes in the closed disc) and $u(z) \to 0, |z| \to 1$ which means by reflection that $u$ is extendable to a harmonic function on some annulus containing the circle $r <|z| <R, R>1$, but now $u=\Re F + c\log z$ for some analytic $F$ and some $c$ on the larger annulus and by analytic continuation $F=f$ on the original part inside the unit disc and $c=0$ hence $f$ extends to an analytic function on the larger annulus so is continuous on the closed unit disc; but now clearly we can extend $f$ to the whole plane as a meromorphic function with finitely many poles, while $\infty$ is a pole or removable depending whether $f(0)=0, \ne 0$ by the usual $f(w)=\frac{1}{\bar f(1/\bar w)}, |w| >1$ and then since $\infty$ is not essential singualarity it immediately follows that $f$ is rational so done!
