Given $A = \{z \in \mathbb{C}: \Im(e^{z}) > 0\}$, describe what the set looks like. Note: $\Im(z)$ is the imaginary part of the complex number $z$.
So if $z = x+iy$, then according to what we were given, $\Im{(e^z)} = \Im{(e^{x+iy})} = e^y >0$. But what does this mean graphically? Does it cover the entire complex-real plane?
If $z = x + iy$, then $$ e^z = e^x \, e^{iy} = e^x (\cos y + i \sin y), $$ so $\Im(e^z) = e^x \sin y$. Since $e^x > 0$ for all $x \in \mathbb{R}$, the upper half-plane condition that defines your set $A$ is equivalent to $$ \sin y > 0. $$
As sine is a periodic function, the inequality is satisfied on the union of open intervals $$ \cdots (-2\pi, -\pi) \cup (0, \pi) \cup (2\pi, 3\pi) \cup \cdots $$ or $$ \bigcup_{n \in \mathbb{Z}} \bigl( 2n\pi, (2n+1)\pi \bigr) $$ and so $A$ is the union of horizontal strips with open boundaries \begin{align} A &= \mathbb{R} \times \bigcup_{n \in \mathbb{Z}} \bigl( 2n\pi, (2n+1)\pi \bigr) \\ &= \bigl\{x + iy \in \mathbb{C} \mid 2n\pi < y < (2n+1)\pi \text{ for some } n \in \mathbb{Z} \bigr\}. \end{align}
