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$f(z)$ = $e^z$, 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 !

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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.

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    great ! so essentially I have a quarter of a circle on the complex plane ? And what about the set builder notation? since I'm expressing it for f(S) would it be f(s) = { S in C such that 0<= I'm(z) <= pi/2} ? – 2017-01-24
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    @NickGuida : If $x$ is fixed at just one number then you have a quarter circle, and you have different points on that arc for different values of $y$. If $x$ is fixed at a different number then you have a different quarter circle, with a different radius. But $x$ doesn't remain fixed: $x$ can be any real number. $\qquad$ – 2017-01-24