$f(z)$ = $e^z$, and $S = \{z\in \Bbb C : 0 \le \operatorname{Im}(z) \le \pi/2 \}$
For $S\subseteq \Bbb C$, and $f$ a complex function whose domain contains $S$.
Sketch $f(S)$ and express it with set builder notation.
I understand that the set $f(S)$ is the image of $S$, and is the set where all $z$ are in $S$. From here I'm not quite sure what to do. I think it would be smart to graph $e^z$ on the complex plane which is a unit circle (?) but only the upper half about the real axis.
In general very confused, please help. Thanks !
Suppose $z = x+iy$ and $x$ and $y$ are real. Then $e^z = e^{x+iy} = e^x(\cos y + i\sin y).$ As $x$ runs through the whole set of real numbers, $e^x$ runs through the whole set of positive numbers. But $y$ goes only from $0$ to $\pi/2$, so you stay in the first quadrant.
You won't have a graph of the function, but only a graph of the image of the function.