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.
 A: 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 $n = 2$.
A: By Cauchy integral formula around a circle centered at $0$ of radius $R$ we have $f^{(n)}(0)=\frac{n!}{2\pi i}\int \frac{f(z)}{z^{n+1}}dz$, taking modulus you have $|f^{(n)}(0)|\leq \frac{n!}{2\pi}\int |\frac{f(z)}{z^{n+1}}|d|z|=\frac{n!}{2\pi}\int \frac{|f(z)|}{R^{n+1}}d|z|\leq \frac{n!}{2\pi R^{n+1}}\int (1+R)d|z|=\frac{n!(1+R)2\pi R}{2\pi R^{n+1}}$ which tends to $0$ as $R\rightarrow \infty$ for $n\geq 2$. Since $f$ is entire we may consider the taylor series expansion centered in zero, the relations between the coefficient of the taylor series and the derivatives of $f$ let us conclude that $f(z)=a_0+a_1z$, which is what we wanted.
A: Here is another approach, in case you're interested. Since $f$ is entire, we can write $f(z)=\sum_{n=0}^{\infty}a_nz^n,$ the power series converging for all $z.$ Now for any $r>0,$
$$2\pi (1+r)^2 \ge \int_0^{2\pi} |f(re^{it})|^2\,dt = \int_0^{2\pi} |\sum_{n=0}^{\infty}a_nr^ne^{int}|^2\,dt = 2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}.$$
For this inequality to hold for all $r,$ we must have $a_n = 0$ for $n>1.$ Thus $f(z) = a_0 + a_1z$ as desired.
