If $D = \{1 + bi\mid b \in\Bbb R\}$ and $f(z) = e^z$, calculate $f(D)$ and describe the image set.

Question

If $D = \{1 + bi\mid b \in\Bbb R\}$ and $f(z) = e^z$, how do you calculate $f(D)$ and describe the image set?

I am thinking $D$ looks like a circle on the complex plane because $D =\{1 + bi\mid b \in \Bbb R\}$, but I don't know how to work on $f(D) = e^D$. Any ideas?

Answer 1

Actually, $D$ is a line not a circle. However, $f(D)$ is a circle, since$$(\forall b\in\mathbb{R}):e^{1+bi}=e(\cos b+i\sin b).$$So, it's the circle centered at $0$ with radius $e$.

Answer 2

$D$ is the line through $1$ and parallel to the imaginary axis.

For $1+ib \in D$, we have $f(1+ib)=ee^{ib}$, hence$|f(1+ib)|=e$, thus $f(D)$ is the circle centered at $0$ with radius $e$.