I have made a sketch of the set $S = \{z \in \mathbb{C} : |z − 3 − 2i| \lt 4 , \operatorname{Re}(z) \gt 0 \}$ on an Argand diagram and the region described by $S$ has no boundary points, since the inequalities are all strict inequalities and the region is described by them.
But the solutions I have to this question describe the set as not open since 'the points on the y axis are not interior points'?
I am not sure whether I am incorrect or not?
On
That set is open, since it is the intersection of two open sets:$$\{z\in\mathbb{C}\,|\,|z-3-2i|<4\}\tag1$$and$$\{z\in\mathbb{C}\mid \,\operatorname{Re}z>0\}.\tag2$$Each of them is open since each of them is of the form $f^{-1}(A)$ for some open set $A\subset\mathbb R$ and some continuous function $f\colon\mathbb{C}\longrightarrow\mathbb R$ (in the case of $(1)$, $f(z)=|z-3-2i|$ and $A=(-\infty,4)$; in the case of $(2)$, $f=\operatorname{Re}$ and $A=(0,+\infty)$).
On
Complex plane:
1) $|z-(3+2i)| <4 $;
$S_1:= B_r(C)$, an open ball with centre $C (3+2i)$ and radius $r=4$.
Hence $S_1$ is open .
2)$S_2:= ${$z|$ $R$e $z >0$} is an open set.
(For $p \in S_2$ , $R$e $p > 0$; Choose
$r= (R$e $p )/2$, then $B_r(p) \subset S_2$)
3) The Intersection $S_1\cap S_2$ of the open sets $S_1,S_2 $ is open.
No points on the $y$-axis are in $S$, so while it's correct that they are not interior it's irrelevant. You're absolutely right in calling the set open.