How to show that $\exp(\mathbb{C}) $ is open subset of $\mathbb{C}^*$, the set of non zero complex numbers How to show that $\exp(\mathbb{C}) $ is open subset of $\mathbb{C}^*$, the set of non zero complex numbers?
The Hint given to me says that, for $w\in \exp({\mathbb{C}})$
$D_{|w|}(w) = wD_1(1)\subset w\exp(\mathbb{C})= \exp(\mathbb{C})$
I don't understand how we got $D_{|w|}(w) = wD_1(1)$
Also, how is $wD_1(1) \subset w\exp(\mathbb{C})?$
 A: I can answer the question.  However, I am unfamiliar with the syntax that the query
is using.  Therefore, I have to use my own syntax.
Let the set $S \equiv ~$ exp$(\mathbb{C})$.
I have to show three things.
(1) $~0 ~\not\in ~S.$
(2) $~z \neq 0 \implies ~z \,\in \,S.$
(3) $~\forall ~z ~\in ~S,~$ there exists a neighborhood $N$ around $z$ such that
$N \subseteq S.$

For any $~(x + iy) ~\in \mathbb{C},$ and
for any $~z ~\in \mathbb{C}$ 
$e^{(x + iy)} = z \iff \{x = \text{ln}|z| ~\text{and}
~y \equiv \text{Arg}(z) \pmod{2\pi} ~\}.$
(1)
As $z \to 0,~ $ ln$|z| \to -\infty.$
For any $~(x + iy) ~\in ~\mathbb{C}, ~x~$ is finite. 
Therefore, for any $~(x + iy) ~\in ~\mathbb{C}, ~e^{(x + iy)} \neq 0.$
(2)
For $z \neq 0,~$ let $~x = \text{ln}|z|~$ and let $~y = \text{Arg}(z).$ 
Then $~(x + iy) \,\in \,\mathbb{C}~$ and $~e^{(x + iy)} = z.$ 
Therefore $~z \,\in \,S.$
(3)
Given any $~z \,\in \,S,~$ from (1) above, $~z \neq 0.$ 
Thus: 
Given any $~z \,\in \,S,~$
define the neighborhood $N$ to be the open circle of radius
$~\frac{|z|}{2}~$ around $~z$.
Take any $~n \,\in \,N$. 
Then, $~n \neq 0.$
Therefore, by (2) above, $~n ~\in ~S.$
