I have been briefly looking into rational functions. The following 2 examples seem obvious, although I cannot think of a concrete explanation:
$e^{(1+z/3)}$, $cos(z+1/z^2)$ are not rational functions
I am assuming this follows because $e^z$ and $cos(z)$ are not rational functions!? But why is this true?
On
There are many possible answers here.
Here is one hint: if $f(z)=\frac{p(z)}{q(z)}$, where $p,q$ are polynomials, then $f(z)$ can have at most finitely many zeros, or it is zero. This comes from the fact that the zeros of $f$ are the zeros of $p$ that are not zeros of $q$, and there are at most $\deg p$ of them.
A consequence: a rational function can have each value $C$ for at most finitely many $z$, or $f$ is equal to a constant $C$. You prove that by applying the statement above to the (rational) function $f-C$.
Now... For how many $z$ is $\cos(z+\frac{1}{z^2})=0$, i.e. can you solve equations $z+\frac{1}{z^2}=\pi/2+k\pi$ for every $k\in\mathbb Z$? (Yes, you can, because those equations reduce to cubic equations in $\mathbb C$, which always have solutions.)
Also... For how many values $z$ is $e^{1+\frac{z}{3}}=1$? (Solve the equations $1+\frac{z}{3}=2k\pi i$ for $k\in\mathbb Z$).
Look at the behaviour on the reals:
The second one is not constant yet bounded, impossible for a rational function. Alternatively, it vanishes infinitely many times on $\Bbb R$, also impossible.
The first one increases at exponential speed, when rational functions should be $O(x^k)$ for some $k\in \Bbb Z$.