$f$ analytic in $\overline{\mathbb{D}}$ s.t $\int_{|z|=1}|f(z)-z||dz|\leq\frac{1}{100}$. Prove that $f$ has at least one zero in $\mathbb{D}$.
Solution 1:
Let $f(z)=\sum a_kz^k$.
Note that $\int_{|z|=1}|f(z)-z||dz|=\int_{|z|=1}|f(z)/z^k-1/z^{k-1}||dz| \ge$
$|\int_{|z|=1}(f(z)/z^k-1/z^{k-1})dz|$,
and this last integral is $2\pi |a_{k-1}|$ if $k \ge 1, k \ne 2$ and $2\pi |a_1-1|$ if $k=2$, so one gets $|a_k| \le 1/(200 \pi)$ if $k \ne 1$ and $|a_1| \ge 1-1/(200\pi)$.
But now one can apply Rouche on say $|z|=1/2$ to $f, a_1z$ as a simple majorization shows that $|f-a_1z| <|a_1|/2$ for $|z|=1/2$ and we are done; one can of course do much better than $1/2$ in finding a $c<1$ st $f$ has a zero with $|z|<c$ if needed