Is the function $f : \mathbb C^{*} \rightarrow \mathbb C^{*} $ given by $f(z)=ze^{z}$ a closed map? I came across a problem where I was asked to find that, the same map is closed or not when seen as from $\mathbb C^{*}$ to $\mathbb C$. So i solved it by showing that image of real line is not closed in the later.
After then I was thinking about the closeness of the same function with co-domain non-zero complex number, but I wasn't able to do it as in the last case zero was playing a crucial role.
Any hint please?
 A: The answer is no and one can show it as follows:
First we note that $ze^{z}=-1$ iff $z+e^{-z}=0$
Let $g(z)=z+e^{-z}$, hence $g$ is entire of order $1$ so it has infinitely many zeroes as otherwise, it would be of the form $P(z)e^{h(z)}$ for some polynomial $P$ and entire $h$ which easily leads to a contradiction by looking at various limits at infinity (eg letting $z=-R \to -\infty$ implies $P$ constant, $h$ linear etc).
Let $|z_1| \le |z_2| \le..\le |z_n| \le ...$ the roots of $g$ and let $w_n=z_n-a_n$ where $a_n >0, a_n \to 0$ st $w_n \ne z_k$ for any $k$; then $|w_n| \to \infty$ and $e^{w_n}+w_n \ne 0$ so $w_ne^{w_n} \ne -1$; however $w_ne^{w_n}=(z_n-a_n)e^{z_n-a_n}=-e^{-a_n}+\frac{a_ne^{-a_n}}{z_n} \to -1$ (using that $e^{z_n}=-1/z_n$ for the second expression), which shows that the image of the closed set (in $\mathbb C^*$) $w_n$ is not a closed set in $\mathbb C^*$
Note that there is nothing special about $ze^z, \mathbb C^*$ etc here as any entire function $f$ that is not a polynomial has all but at most one value as asymptotic value (in other words for all but at most one $c$ there are $|z_n| \to \infty, f(z_n)=c$) and then choosing small perturbations of the $z_n$ as above gives a set $w_n, |w_n| \to \infty$ for which $f(w_n) \ne c, f(w_n) \to c$, showing that the image of the closed plane set $w_n$ is not closed in the plane, plane minus any number of points etc)
