Showing that $f$ is linear function if $\forall z \in \mathbb{C}$, $|f(z)| \leq 1 + |z|$. Let $f$ be an entire function that satisfies $|f(z)| \leq 1 + |z|$ for all $z\in \mathbb{C}$. Show that $f(z) = az +b$ for fixed complex numbers $a,b$. The hint tells us to try and use Cauchy's Integral Formula on an arbitrary circle. This is my attempt: Consider an arbitrary large circle with centre $0$ and let $\Upsilon$ be the contour around this circle. Then $$\int_{\Upsilon} \frac{f(z)}{z^{n+1}} \mathrm{d}z = \frac{2\pi i}{n!}f^{(n)}(0).$$ Note that $$\int_{\Upsilon}\frac{f(z)}{z^{n+1}} \leq \int_{\Upsilon}\left|\frac{f(z)}{z^{n+1}} \right|\leq \int_{\Upsilon} \frac{1}{|z^{n+1}|} + \frac{1}{|z|^n}$$ and the right hand side is $0$ by Cauchy's Integral Formula. I'm not sure if the first equality is valid, and not sure where to go from here.

Here's the big missing idea: if a function's second derivative is identically zero, then the function is at most a linear polynomial. So you should show that the right hand side is zero when $n = 2$.