exponential function never vanishes

In the entry exponential function (http://planetmath.org/ExponentialFunction) one defines for real variable $x$ the real exponential function  $\exp x$, i.e. $e^x$, as the sum of power series:

$$e^x=\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}$$

The series form implies immediately that the real exponentialfunction attains only positive values when  $x\geqq 0$.  Alsofor    the positiveness is easy to see bygrouping the series terms pairwise.

In to study the sign of $e^x$ forarbitrary real $x$, we may multiply the series of $e^x$ and$e^{-x}$ using Abel’s multiplication rule for series (http://planetmath.org/AbelsMultiplicationRuleForSeries).  We obtain

$$e^x{e}^{-x}=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k\frac{x^k}{k!}=\sum _{n=0}^{\mathrm{\infty }}\sum _{j=0}^n{(-1)}^j\frac{x^n}{j!(n-j)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right){(-1)}^j=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\cdot 0^n.$$

The last sum equals 1.  So, if  $-x>0$,  then  $e^{-x}>0$,  whence $e^x$ must be positive.

Let us now consider arbitrary complex value  $z=x+iy$  where $x$ and $y$ are real.  Using the addition formula of complex exponential function and the Euler relation, we can write

$$e^z=e^{x+iy}=e^x{e}^{iy}=e^x(\cos y+i\sin y).$$

From this we see that the absolute value of $e^z$ is $e^x$, which we above have proved to be positive.  Accordingly, we may write the

Theorem.  The complex exponential function never vanishes.