If $|f'(z)| \leq |z|$ for all $z\in \mathbb{C}$ then $f(z) = a + bz^2$ for arbitrary $a,b\in\mathbb{C}$ with $|b| \leq 1$. Is there something wrong with this problem?

Let $f$ be entire function and $|f'(z)| \leq |z|$ for all $z\in
\mathbb{C}$. Show that $f(z) = a + bz^2$ for arbitrary
  $a,b\in\mathbb{C}$ with $|b| \leq 1$.

I was using Liouville theorem, but the result is not the same. 
By writing $$\left|\frac{f'(z)}{z}\right| \leq 1$$ we have that $f'(z)/z$ (*here where I'm not sure) is entire and bounded by $1$. So by Liouville's theorem $f'(z)/z$ is constant, that is $$f'(z)/z = c$$ hence $f'(z) = cz$ and therefore $f(z) = \frac{1}{2}cz^2 + a$. Letting $\frac{1}{2}c = b$ yields the result, but with $|b| = \frac{1}{2}|c| \leq \frac{1}{2}\times1 = \frac{1}{2}$ instead of $|b| \leq 1$.
Here are the questions


*

**How to show that $f'(z)/z$ also analytic at $z = 0$?

*Is there an error in my proof above? Or with the problem?


If $f'(z)/z$ need not be analytic at $z = 0$ then I can't use Liouville theorem. All I know is that $f'(0) = 0$, so I think that $f'(z)/z$ is also analytic at $z = 0$.
 A: You know that your function is entire because 
$$
|\frac{f'(z)}{z}| \leq 1
$$
is bounded. Since it is bounded and holomorphic around 0, you can extend it to $z=0$ and therefore you have an entire function. It is a "removable singularity", you can find characterisations of it in any textbook or script.
As for your other question:
It is trivially true that $|b|\leq \frac{1}{2}\leq 1$. You can just plug in the derivative and check that 
$|bz| \leq |z|$ implies $|b| \leq 1$. In your proof, you had a statement on the derivate, i.e. $f'(z)=cz$. And then you integrated both sides. In the process, you got a factor of $\frac{1}{2}$ which, however, vanishes when you differentiate with respect to $z$.
The problem here is that you take a statement/restriction on the derivative and by integrating the statement changed. Look at it that way, just ignore the $|.|$ for a second:
$$
f'(z) \leq cz
$$ 
becomes:
$$
\int f'(z)=f(z)\leq  \frac{1}{2}cz^2=\int cz
$$
A: Let $z_0\in\mathbb{C}$ be arbitrary. Then for $|z-z_0|\leq R$, we have
$$|f'(z)|\leq |z| = |z-z_0+z_0|\leq R+|z_0|.$$
By cauchy's inequality, 
$$\left|\frac{f'''(z_0)}{2!}\right|\leq \frac{R+|z_0|}{R^2}\rightarrow 0 \mbox{ as } R\rightarrow \infty.$$
So, $f'''(z_0)=0$. Since $z_0$ was arbitrary, $f'''(z)=0$ on $\mathbb{C}$. Now, integrate it from $0$ to $z$, we get 
$$f''(z)=f''(0).$$
Again, integrate from $0$ to $z$,
$$f'(z)=f''(0)z+f'(0).$$
As $f'(0)=0$, $f'(z)=f''(0)z$.
Therefore, $f(z)=a+\frac{f''(0)}{2}z^2$ where $a$ is arbitrary. Take $b=\frac{f''(0)}{2}$. So, $|b|=|\frac{f''(0)}{2}|\leq 1$, as $|f''(0)|\leq 1$.
Observe that, $f''(0)=\lim\limits_{z\rightarrow 0}\frac{f'(z)}{z}$, as $f'(0)=0$.
