Entire Function Bounded By Linear Function

Since $f : \mathbb{C} \to \mathbb{C}$ is a holomorphic function and $|f(z)| \leq 5|z| \Rightarrow f(0) = 0$ , the power series of $f$ around $z = 0$ is : $f(z) = \displaystyle \sum_{k = 1}^{\infty} \frac{f^{k)}(z)}{k!} z^k \Rightarrow$ the function $g: \mathbb{C} \to \mathbb{C}$ such that $g(z) = \frac{f(z)}{z}, \space \forall z \in \mathbb{C}$ is a bounded holomorphic function and due to Liouville's theorem $g(z) = C$ whith $C$ a constant. This implies that , $g(z) = \frac{f(1)}{1} = 3 + 4i, \space \forall z \in \mathbb{C} \Rightarrow f(1 + i) = (3 + 4i)(1 + i) = - 1 + 7i$

from the equation we see that

$\left|\frac{f(z)}{z}\right|\le 5$

By Liouville's amazing theorem this means that $\frac{f(z)}{z}$ is constant. So substituting in the values given in the problem gives

$\frac{3+4i}{1}=\frac{f(i+1)}{i+1}$

Then easy algebra gives

$f(i+1)=7i-1$