meromorphic function in the unit disc with only one pole of order n Let $f$ be meromorphic in a neighborhood of $\{|z| \leq 1\}\setminus \{1/2\}$ and have a pole or order $n$ at $1/2$. Suppose that $|f| < 3$ on $\{|z|=1\}$. Show that for any $\phi \in \mathbb{R}$, $f$ attains the value $3e^{i\phi}$ exactly $n$ times (counting multiplicities) in $\{|z| \leq 1\}$.
I suspect that somehow Rouche's theorem will be needed to complete this proof. I started with observing that there exists a holomorphic function $g(z)$ in $\mathbb{\bar{D}}$ given by $g(z) = (z-1/2)^n f(z)$ such that $g(1/2) \neq 0$. Therefore using the maximum principle I can write that in $\mathbb{\bar{D}}$, $|g| < \max_{|z|=1} |z-1/2|^n |f| < 3\dfrac{3^{n}}{2^n}$. But I am confused on what holomorphic functions to choose and compare while applying Rouche's theorem. 
 A: Take $u(z) = - 3 e^{i \phi} (z- \frac 1 2)^n$. Then $u(z)$ and $g(z)$ both are holomorphic inside the closed unit disk. In the unit circle, $|u(z)| = 3 |z-\frac1 2|^n > |f(z)| |z- \frac  1 2|^n = |g(z)|$. So by Rouche's theorem number of roots of $g(z) - 3^{i \phi} (z - \frac  12)^n$ and $3^{i \phi} ( z - \frac 12)^n$ inside $|z| \le 1$ are equal, that is $n$ and those are precisely the solutions of $f(z) = 3 e^{i \phi}$.
A: I can show that we have a lower bound of $n$.
Let $\mathcal{C}$ denote the set of solutions inside the unit disk to the equation $|f|=3$. Since we have a pole at $1/2$, there will be a closed curve around $1/2$ where $f$ has modulus $3$ - indeed for each real angle $\theta\in[0,2\pi)$, follow the line segment with angle $\theta$ originating at $1/2$ until we reach a point $z$ such that $|f(z)|=3$. We must reach such a point since $|f|<3$ on the unit disk, and the union of all of these points will form a closed curve due to continuity of $f$. Combining the fact that $|f|<3$ on the unit circle with the maximum modulus principle, we can conclude that $\mathcal{C}$ consists only of this closed curve around the pole.
Applying the argument principle, we have that $$\int_{\mathcal{C}} \frac{f'(z)}{f(z)}dz =-2\pi n i,$$ and so the change in argument around the circle is precisely $-2\pi ni$. This implies that for every real number $\phi$, the function $f$ attains the value $3e^{i\phi}$ at least $n$ times inside the unit disk.
