I'm trying to find out what is the image of the group $L=\{z\in\Bbb C\mid a\operatorname{Re}(z)+b \operatorname{Im}(z)+c=0\,; a^2+b^2\neq0\,; a,b,c\in \Bbb R \}$ under the function $e^z$ in the Imaginary-real plane. In other words, find $\{e^z\mid z\in L\}$.
Correct me if something is not good in the question.
I understand it should be something like a spiral, but I don't really understand how to show it.
Hint: $\,a,b\,$ cannot both be zero. Suppose for example that $\,a \ne 0\,$, then it can be assumed WLOG that $\,a=1\,$ (by dividing with $\,a\,$ and redefining $\,b \mapsto b/a$, $\,c \mapsto c/a\,$).
Let $\,t=-\operatorname{Im}(z) \in \mathbb{R}\,$, then the equation gives $\,\operatorname{Re}(z) = bt -c\,$. It follows that $\,L\,$ can be parameterized $\,z = bt-c-it\,$ with $\,t \in \mathbb{R}\,$, and therefore $\,w = e^z=e^{bt-c}\cdot e^{-it}\,$. Note that the modulus $\,|w| = e^{bt-c}\,$ varies monotonically with $\,t\,$ (increases or decreases depending on the sign of $\,b$), and its argument $\,\arg w = -t\,$ decreases with $\,t\,$, which describes a spiral (except for the case $\,b=0\,$ when it degenerates into a circle).