Prove that the range of the entire function $z^2+\cos(z)$ is all of $\mathbb{C}$. 
Prove that the range of the entire function $z^2+\cos(z)$ is all of $\mathbb{C}$.

I'm aware this question has been asked already, but the explanations were a little shakey and referenced a google sample preview of a text of whose terminology I wasn't very familiar with, so I would like some more input if possible.
Here's my attempt:
Let $f(z)=z^2+\cos(z)$. Through the power series of the cosine function
$$
f(z)=z^2+1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots=1+\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots,
$$
so
$$
g(z):=\frac{f(z)-1}{z}
$$
has a removable singularity at $z=0$ and is effectively an entire odd function. Picard's little theorem asserts that the range of $g$ is all of $\mathbb{C}
$: if there exists some $z_0$ such that $g(z)\neq z_0$ for all $z\in\mathbb{C}$, then $-z_0$ is also missed by oddness of $g$, which is a contradiction (since $g(0)=0$ we rule the possibility that $z_0=0$). (I'm not sure about the following step) It follows that $zg(z)$ and so $zg(z)+1=f(z)$ is surjective.
Is this method valid? Would I need further justification to show that $z\mapsto z$ and $z\mapsto g(z)$ surjective imply its product -- $z\mapsto zg(z)$ -- is also surjective (if this is even true). Any input is greatly appreciated.
Edit: After thinking about this for a moment longer, I realize this logic would also show that the range of any entire even function is all of $\mathbb{C}$, which seems like quite a strong conclusion. This makes feel like there are holes in my logic.
 A: No:  there's no reason for $g$ surjective to imply that $z g(z)$ is surjective.
Thus $g(z) = (e^{z^2}-1)/z$ (removing the removable singularity at $0$) is odd and entire, therefore surjective, but $z g(z) = e^{z^2}-1$ does not take the value  $-1$.
EDIT: In your problem, with $f(z) = z^2 + \cos(z)$, any missing value of $f(z)$ must be real ($f(\overline{z}) = \overline{f(z)}$, so if $w$ is missing so is $\overline{w}$, but there can be only one missing value by Picard's theorem).  Moreover, real $z$ gives you all values on  $[1,\infty)$,
so it suffices to consider $w \in (-\infty, 1)$.
Now I can use the symmetric version of Rouché's theorem to show that $f(z)-w$ has the same number of solutions as $g(z) - w= \cos(z) - w$ (and thus at least one)
inside a rectangular path
$\Gamma$ with corners $\pm 2 n \pi \pm n i$ where $n$ is large.  
On the horizontal segments of this path, $z = t \pm n i$ for $-2\pi n \le t \le 2\pi n$,
$|f(z) - g(z)| = |z^2| \le (4 \pi^2 + 1) n^2$
while $|\cos(z)| \ge \sinh(n)$, so $| g(z) - w| \ge \sinh(n) - |w| > |f(z) - g(x)|$ if $n$ is large enough.  
On the vertical segments, $z = \pm 2\pi n + i t$, $-n \le t \le n$, $\cos(z) = \cosh(t) > w$ and $\text{Re}(z^2) = 4 \pi^2 n^2 - t^2 \ge (4 \pi^2 - 1) n^2 > w$ if $n$ is large enough.
Now the only way to have $|f(z) - g(z)| \ge |f(z)-w| + |g(z)-w|$ is that 
$f(z) - w$ and $g(z) - w$ are on opposite sides of a line segment passing through $w$, and in particular $\text{Re}(f(z)-w)\le 0$ or $\text{Re}(g(z) - w \le 0$, so that can't happen.
