A Question on the Application of Liouville's Theorem Let $f(z)$ be an entire function that satisfies $|f(z)|\leq 1+|z|, \ \forall z\in\mathbb{C}$. Show that $f(z)=az+b$ for fixed complex numbers $a$ and $b$.


My attempt:

Consider a circle, $\Gamma$, centered at the origin with radius $R$. Now, by Cauchy's Integral Formula:
$$f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \ \ \forall z,$$
where, $$f^{(n)}(0)=\frac{n!}{2\pi i}\int_\Gamma \frac{f(z)}{z^{n+1}} \ dz.$$
Now suppose $\Gamma=\Gamma_R=Re^{it} \ \ t\in[0,2\pi]$. Then if $|z|=R$, we have:
$$\left|\frac{f(z)}{z^{n+1}}\right|\leq\frac{1+R}{R^{n+1}}.$$
So, $$|f^{(n)}(0)|=\frac{n!}{2\pi}\left|\int_{\Gamma_R} \frac{f(z)}{z^{n+1}} \ dz\right|\leq\frac{n!}{2\pi}\frac{1+R}{R^{n+1}}2\pi R=\frac{n!(1+R)}{R^n}.$$
Now $|f^{(n)}(0)|\rightarrow 0$ as $R\rightarrow\infty$ for $n\geq 2$. But what is the answer (what is $a$ and $b$)?
What you have proved is  $f^{(n)}(0)=0$ for $n \geq 2$ because $f^{(n)}(0)$ does not depend on $R$. From the power series expansion you get $f(z)=f(0)+f'(0)z$ so $a=f'(0), b=f(0)$.
Each pair $(a,b) \in \mathbb C^2$ with $|a|,|b| \le 1$ will do the job !