Find entire functions that satisfy the following inequalities I am trying to determine whether there exist two entire functions $f(z),g(z):\mathbb{C}\to\mathbb{C}$ that satisy the following inequalities (independently): 

$$(1): |e^{|z|}-f(z)|&lt;e^{-|z|}\\(2):|g(\bar{z})|\geq e^{|z|}$$

Well, as for inequality $(1)$, I figured that if there exists an entire function $f(z)$, it has to be some sort of an exponent. A (finite) polynom wouldn't work, because the LHS of the inequality would always be unbounded (whereas the RHS is indeed bounded). I tried substituting $f(z)=e^{az}$ for some $a\in\mathbb{C}$ but couldn't figure out what to do further.

Same goes for inequality $(2)$. Whereas I need the inequality to be true for every $z\in\mathbb{C}$, no matter what function I plug in, I find out that some sort of a connection between $\Re{(z)}$ and $\Im{(z)}$ must happen for the inequality to hold.

Those consequences led me to think that maybe there don't exist entire functions as required, but I couldn't prove why is this true. Something tells me that I need to use Liouville's Theorem, but I couldn't find the connection. I tried, for example, in equality $(1)$, to multiply both sides by $e^{|z|}$. Then, the new LHS is bounded. I wanted to say it is constant because of Liouville's Theorem, but the problem is that the LHS I get is not an entire function (thus I cannot use the theorem); neither in inequality $(2).$ Of course there's a chance that I'm wrong and just don't know enough entire functions in order to find a good example.

I'm pretty lost right now so I'd be happy to hear your thoughts. Thanks!
(1) would imply that $f$ is non-zero on the whole complex plane, then the function $1/f$ would be entire and bounded, hence constant. This is not possible.

For (2) it's the same trick: consider $1/g$, which must be entire and bounded.
There are no such entire functions. Since a polynomial has sub-exponential growth, this is immediate from the following (which I bet was a previous result or a previous exercise in the book):


  
    Lemma. If $f$ is an entire function and $\lim_{|z|\to\infty}|f(z)|=\infty$ then $f$ is a polynomial.
  


Indeed. say $$f(z)=\sum_0^\infty c_nz^n.$$Then $$f(1/z)=\sum_{-\infty}^0c_{-n}z^n.$$

But $\lim_{z\to0}|f(1/z)|=\infty$, which bt definition says $f(1/z)$ has a pole at the origin. Looking back at the section on Laurent series for functions with various sorts of singularities, we see this implies $$f(1/z)=\sum_{-N}^\infty a_nz^n.$$So uniqueness of the Laurent series implies that $$c_n=0\quad(n&gt;N).$$