Prove that $e^{-a}=1/e^a$ I'm preparing for a test and I realized that I have no idea on how prove an inverse statement, such as $e^{-a}=1/e^a$? Furthermore if anyone can provide hints on how to do this for complex numbers, $x+iy$ that would be great. I've looked around for anything to do with complex numbers but most of them involve Laurent's Series.
 A: How you prove this depends on how you defined $e^x$. The simplest way is by the power series: $$\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}.$$
It's easy to from that that $\exp'=\exp$. Now fix $y$ and define $$f(x)=\exp(x+y)\exp(-x).$$It follows from the product rule that $f'=0$, so $f$ is constant. In particular $$\exp(y)=f(0)=f(x)=\exp(x+y)\exp(-x)$$.
So $$\exp(a+b)=\exp((a+b)-b)\exp(b)=\exp(a)\exp(b).$$
So $$1=\exp(0)=\exp(x)\exp(-x).$$
Exactly the same argument works for complex numbers, if you know what it means for $f:\mathbb C\to\mathbb C$ to be differentiable.
A: Note that 
$$e^a=e^{x+iy}=e^xe^{iy}=e^x(\cos y+i \sin y)$$
$$e^{-a}=e^{-x-iy}=e^{-x}e^{-iy}=e^{-x}(\cos y-i \sin y)$$
thus
$$e^a \cdot e^{-a}=e^{x-x}(\cos y+i \sin y)(\cos y-i \sin y)=1$$
A: It depends on your definition of $e^x$.
(Remember this! In mathematics, everything must have a formal definition.)
Let's assume you have formally defined exponentiation by
$$
e^x := 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.
$$
(Or, it is enough that you have proved this equality as a lemma.) Then,
\begin{align*}
e^{-a}
&= 1 - a  + \frac{a^2}{2!} - \frac{a^3}{3!} + \cdots \\
e^a
&= 1 + a  + \frac{a^2}{2!} + \frac{a^3}{3!} + \cdots
\end{align*}
Here is our plan to show that $e^a e^{-a}$ is identically $1$: first, show that it's derivative is $0$, so it is constant; then show, that constant is $1$.


*

*Derivative is 0 We know the derivative of $e^a$ and $e^{-a}$ by differentiating the power series. So we use the product rule:
$$
  \frac{d}{da} (e^{a} e^{-a}) = e^a (- e^{-a}) + (e^a) (e^{-a}) = 0.
  $$

*The constant must be 1 Just plug in $a = 0$. Then we get $e^0 e^0 = 1$.
In summary: since $e^a e^{-a}$ is an entire function (radius of convergence $\infty$!), its derivative is everywhere $0$, and it takes on the value $1$, it must be the constant function $1$. Thus, $e^a e^{-1} = 1$ for all $a$. And this reasoning works the same in the real and the complex case.
