Let $z = a + b*i$ where we assume a and b are real numbers. Prove $e^{-z} = \frac{1}{e^{z}}$

Question

What is the easiest way to prove that $e^{-z} = \frac{1}{e^{z}}$ when $z = a + b*i$ where we assume a and b are real numbers.

Answer 1

There are (at least) three options, depending on how you're defining $e^z$:

(1) If you define $e^{a + bi} = e^a (\cos b + i \sin b)$, then it follows just from algebraic manipulation and the corresponding real result.

(2) If you define $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$, then the result follows with some more algebraic manipulation, probably along with some sort of result about uniform convergence on compact sets.

(3) A better definition is to take $e^z$ to be the unique solution of the differential equation $f' = f$ with $f(0) = 1$. Then for fixed $w$, the function $g(z) = e^{z + w} / e^w$ satisfies the same equation and boundary condition, so we have $e^{z + w} = e^z e^w$ everywhere. This approach requires results on the existence and uniqueness of certain differential equations, though.