Is there a function $f$ such that $f$ is entire and for all $z\in\mathbb{C}$ with $|z|\geq 100:|f(z)|=|z|+1$.
An observation is that $\lim\limits_{z\to\infty}f(z)=\infty$ and so $f$ must be a polynomial.
Therefore, the problem reduces to showing that no polynomial can uphold the given constraint.
My problem is in doing so. Is there another way?
Let $a\in\mathbb{C}$. For a large enough $R>100,|a|$, by Cauchy's formula we get
$$f''(a)=\frac{1}{\pi i}\int_{|z|=R}\frac{f(z)}{(z-a)^3}dz\implies $$
$$|f''(a)|\leq\frac{1}{\pi}\frac{1}{|R-a|^3}\cdot2\pi R$$
The latter goes to $0$ as $R\to\infty$, so $f''(a)=0$. Therefore, $f''\equiv 0$.
We get that $f$ is a linear function $f(z)=az+b$.