TITLE: Prove that the exponential $\exp z$ is not zero for any $z \in \Bbb C$
QUESTION [6 upvotes]: How can the following been proved?

$$
\exp(z)\neq0, z\in\mathbb{C}
$$

I tried it a few times, but I failed. Probably it is extremly simple.
If a draw the unit circle and then a complex number $\exp(a+ib)=\exp(a)\exp(ib)$ then it is obvious that this expression is only $0$ if $\exp(a)$ equals zero, but $\exp(a),a\in\mathbb{R}$ is never zero. This seems not to be very robust.
Thank you

REPLY [2 votes]: $\exp(z)\neq 0 $ $\space , \forall z\in \Bbb{C}$
Proof:
$f(z) =\exp(z) $ is an entire function i.e holomorphic on $\Bbb{C}$.
To show : $\scr{Z_f}=\{z:f(z)=0\}=\emptyset$
We argue by contradiction. Suppose$\scr{Z_f}\neq\emptyset$.
Let, $z_0\in \scr{Z_f}$
Then, $f^n{(z_0)}=\exp(z_0) =0$ $ \space, \forall n\in \Bbb{N}$
Since, $f\in H(\Bbb{C})$, then $f$ has a Taylor Series  about the point $z_0$ converges for all $z\in \Bbb{C}$.
$f(z) =\sum_{n=0}^{\infty}\frac{f^n(z_0) }{{n!}}(z-z_0)^n$$\space , \forall z\in \Bbb{C}$
Hence, $f(z) =0$ on $\Bbb{C}$.
But, $f(0) =\exp(0) =1$ ,$\space $ a contradiction.
Hence, $\scr{Z_f}=\{z:f(z)=0\}=\emptyset$
In other words, $\exp(z) \neq 0$ for any $z\in \Bbb{C}$