Proof of $e^z \neq 0$

Question

Proof: Let $a \in \mathbb{C}$ be s.t. $e^{a} =0$. Then $0=e^a e^{-a} = e^{-a+a} = e^0 =1$, contradicting the existence of $a$.

But why can we multiply by $e^{-a}$?? If $e^a=0$, then $e^{-a}=\frac{1}{e^a}=\frac{1}{0}$ which can't be defined?

Answer 1

How are you defining $e^a$ over the complex numbers in the first place?

If this is done by the usual series for $$e^x=1+\sum_1^\infty \frac {x^n}{n!}$$ extending the definition from the real numbers, then the series is absolutely convergent for all $x$, then $e^{-a}$ exists and is defined already.

The property $e^xe^{-x}=1$ can be shown from the definition. Then everything is in place to derive a contradiction from the assumption $e^a=0$.

But how you approach the problem depends on how $e^x$ has been defined in the first place.

Answer 2

Your objection is valid. I'd do it in a different way: we know that $$ e^{x+y}=e^x\cdot e^y $$ for all $x,y\in\mathbb{C}$. In particular $$ e^x\cdot e^{-x}=e^{x-x}=e^0=1 $$ and therefore $e^x\ne0$, because $0$ is not invertible in $\mathbb{C}$.

Answer 3

For every complex number $$ z=a+bi $$

we define $$ e^{z}=e^{a}(\cos(b)+i\sin(b)) $$

This expression is well defined for all $a,b\in\mathbb{R}$, hence for all $z\in\mathbb{C}$.

For a different proof you may note that $$ |e^{z}|=|e^{a+bi}|=e^{a} $$

I assume that you know $e^{x}\neq0$ for all real $x$ (in our case we can set $x=a$, as $a$ is real)

Answer 4

The exponential function is defined at all complex arguments; its domain is all of $\bf C$. As such, the value $e^a$ always exists, for all values of $a\in\bf C$, so of course you can multiply by it. Now, one may see that $e^a=0\implies e^{-a}~~``="\,1/0$ is not defined, contradicting the fact that it is, so isn't this reasoning fallacious? No, firstly obtaining a contradiction from $e^a=0$ is kind of the point (it shows that the equality $e^a=0$ is not possible). Secondly you can't actually divide by $0$, which is a step used in obtaining $e^{-a}=1/0$; rather you get $1=e^ae^{-a}=0\cdot e^{-a}$ and instead of dividing by $0$ (which we're not allowed to do) all we can really do is simplify to $1=0$, a contradiction, and of course this also shows that $e^{-a}$ doesn't exist, as $1=0\cdot x$ has no solution.

Ultimately the reason we can multiply by $e^{-a}$ is because we're assuming it exists by hypothesis, which we can do because this is exactly the framework of proof-by-contradiction.