$$S=\{z: -1<Re(z)<2\}$$ $$w=f(z)=e^z$$
Graph $S$ and $f(S)= \{w\}=\{f(z)\mid z \in S\}$
My attempt:
This is what I think it should look like. Correct me if I'm wrong, but I think the general concept is that when graphing $z$ you do it based on any constraints of $x$ and $y$. For $w$ you graph the function $f(z)$ but still keeping in mind the previous constraints of $x$ or $y$. Is this on the right track?
Update: This is what I came up with based on an answer given. If we write $e^z$ in terms of $x$ and $y$, and also limit the $x$ based on the constraint given, then the resulting image of $f(S)$ should be an open disc that does not include $r=1/e$ or $r=e^2$.
The set $S$ is the points of the plane with real part between $-1$ and $2$. I believe your sketch is a perfectly reasonable representation of this set.
The set $f(S)$ is a little different...
Let $s \in S$. You know $\Re(s) \in (-1,2)$ (that's "the real part of s is in $(-1,2)$"). So if we write $s = x + \mathrm{i} y$, we know $-1 < x < 2$. But $y$ can be anything. So let's pick a real part, say $1/2$ and let the imaginary part wave around. We want to know what figure we get from $f(1/2 + \mathrm{i} y) = \mathrm{e}^{1/2 + \mathrm{i} y} = \sqrt{\mathrm{e}} \cdot \mathrm{e}^{\mathrm{i} y}$. The $\sqrt{\mathrm{e}}$ is the magnitude of the point and the $\mathrm{e}^{\mathrm{i} y}$ gives an angle ($\arg$). But that means the line $\Re(s) = 1/2$ gives the circle of radius $\mathrm{e}^{1/2}$.
What does that mean the other choices of real part give us?
What do we get when we glue all those figures together?