Let $f$ be a holomorphic function on a punctured unit disk $D = \{z: 0<|z|<1\}$. If $f$ satisfies the inequality $|f(z)|\le 1+\ln \left( \frac{1+|z|}{2|z|} \right)$ for all $z\in D$.
(a) Prove that $z=0$ is the removable singularity of $f$.
(b) If $f \left(\frac12 \right)=1$, find $f\left(\frac{1+i}{3}\right)$.
My attempt for (a): First I tried to use Cauchy's inequality $|f^{(n)}(z)| \le \frac{n!M}{R^n}$ where $M$ is the maximum of $f$ on the circle centered at $z$ and radius $R$, but it didn't give any information about $f$ in the case $n=0$.
If I solve (a), then (b) would follow directly, I guess, but I don't know how to proceed. Does anyone have any ideas?
$|zf(z)|\leq |z|+|z|\ln(1+|z|)-|z|\ln(2|z|)$
Since $\lim_{r\to0}r\ln(2r)=0$ then $zf(z)$ is bounded near $0$. Therefore $f(z)$ can, at most, have a pole of order $1$. Assume $f(z)=g(z)/z$ with $g$ holomorphic near $0$.
Then $|g(z)|\leq |z|+|z|\ln\left(\frac{1+|z|}{2|z|}\right)$. The same limit as above gives $g(0)=0$. Therefore $f$ doesn't even have a pole of order $1$.