If $f: D(0,1)\to \mathbb C$ is holomorphic, then there is a sequence $\{z_n\}$ in $D(0,1)$ such that $|z_n| \to 1$ and $\{f(z_n)\}$ is bounded

Let $$f: D(0,1)\to \mathbb C$$ be a holomorphic function. How to show that there exists a sequence $$\{z_n\}$$ in $$D(0,1)$$ such that $$|z_n| \to 1$$ and $$\exists M>0$$ such that $$|f(z_n)| ?

My try: If not, then $$\lim_{|z|\to 1} |f(z)|=\infty$$. So in particular, $$f$$ has finitely many zeroes in $$D(0,1)$$. Also, $$1/f$$ is meromorphic in $$D(0,1)$$ with $$\lim _{|z|\to 1}\dfrac {1}{|f(z)|}=0$$. I am not sure where to go from here. Please help.

The way to do it is to first take out the (finitely many) zeros with a finite Blaschke product $$B$$ - which has absolute value $$1$$ on the circle- so get $$g = \frac{f}{B}$$ analytic, no zeros, $$g \to \infty$$ on the circle, so then $$\frac{1}{g}=\frac{B}{f}$$ is an analytic function in the unit disc that goes to zero at the boundary and that means it is zero by maximum modulus, so you get a contradiction and are done.
Let $$w_1,.., w_m$$ be the zeroes of $$f$$ with multiplicity $$k_1,..,k_m$$. Let $$P(z):= (z-w_1)^{k_1}....(z-w_m)^{k_m}$$. Then $$g(z):= \frac{f(z)}{(P(z)}$$ is holomorphic on $$D(0,1)$$ and non-vanishing on $$D(0,1)$$.
Then $$\frac{1}{g(z)}$$ admits a continuous extension to $$\{ z : |z| \leq 1 \}$$, namely the function which is $$0$$ on the boundary.
By compactness, $$|\frac{1}{g(z)}|$$ has a local maximum on this domain. Show that this local maximum appears at some point in $$D(0,1)$$ and use the maximum modulus principle.
• $1/f(z)$ admits an extension towards the boundary, it may not be holomorphic inside ... indeed say, if $f(0)=0$, then $1/f$ is not holomorphic at $0$ – user102248 May 16 at 1:25