Find all entire functions that satisfy following inequality Find all entire functions that satisfy following inequality:

$$ |f(z)| \leq |z| e^{\Re(z)} $$ for all $ z \in \Bbb C $
With

$z = x + iy, \tag 1$

since $\vert e^{iy} \vert = 1$ we have

$e^{\Re(z)} = e^x = \vert e^x \vert = \vert e^x \vert \vert e^{iy} \vert = \vert e^{x + iy} \vert = \vert e^z \vert; \tag 2$

thus

$\vert f(z) \vert \le \vert z \vert e^{\Re(z)} = \vert z \vert \vert e^z \vert = \vert z e^z \vert, \tag 3$

which implies

$f(0) = 0; \tag 4$

it follows then that both $z^{-1}f(z)$ and $e^{-z}z^{-1}f(z)$  are also entire and that (3) yields

$\vert z^{-1} e^{-z} f(z) \vert \le 1, \forall z \in \Bbb C; \tag 5$

therefore, via Liouville's theorem (a bounded entire function is constant),

$z^{-1} e^{-z} f(z) = c \in \Bbb C, \tag 6$

where in fact

$\vert c \vert \le 1, \tag 7$

whence

$f(z) = cze^z, c \in \Bbb C, \; \vert c \vert \le 1. \tag 8$

.
$|\frac{f(z)}{e^z}|\leq|z|\implies f(z)=e^z(az+b)$(by Liouville theorem) $\implies f(z)=e^z(az)$ as $f(0)=0$.