If we have an open set $S$ and multiply every element of $S$ by a fixed complex number $z_{0}$, is the resultant set open? Suppose S in a complex open set with elements $s \in \mathbb{C}$, if we multiply all elements by a fixed complex number $z_0$, will the resultant set be open as well?
My intuition says yes, as multiplying a given open set geometrically has the effect of rotating and stretching the area defined by the set, so this new area should also be open. I am just not sure how I would go about proving it.
Any help would be appreciated.
EDIT: As some have pointed out the constraint on $z_0$ is that it is non-zero.
 A: I think your intuition is in fact a proof. Rotations and stretches (and translations) preserve the topology of the plane. If you need this result to prove something else deeper, just state it and use it.
If you must have something more formal (perhaps for a class exercise) then note that an open ball becomes an ellipse in which you can inscribe an open ball. That will give you the open neighborhoods of points you need.
A: As you were told in the comments, the statement is indeed true if $z_0\ne0$ (and trivially false if $z_0=0$). Take $s\in S$. There is some $\varepsilon>0$ such that the open disk $D_\varepsilon(s)$ is a subset of $S$. But then $D_{|z_0|\varepsilon}(z_0s)$ is a subset of $z_0S$, since it is equal to $\{z_0\times z\mid z\in S\}$. So, $z_0S$ is an open set.
A: If $S \subset \Bbb C$ is open and $z_0 \ne 0$ then
$$
 T = z_0 S = \{ z_0 s \mid s \in S \}
$$
is open as the pre-image of an open set under a continuous mapping:
$$
 T = f^{-1}(S)
$$
with $f(z) = z/z_0$.
If $z_0 = 0$ (and $S$ is not empty) then $T = \{ 0 \}$ is not open.
A: This follows from the open mapping theorem, since for $z_0\neq0,$ the function $f(z)=z_0z$ is nonconstant and analytic. It's analytic because it's holomorphic,  with derivative $f'(z)=z_0.$
Actually,  $z_0z$ is a polynomial,  so already a power series.   So it's clear $f$ is analytic in the first place.
