Proving $e^{-z}=\frac{1}{e^z}$

Question

Prove $$e^{-z}=\frac{1}{e^z}$$.

Taking $z=z_1+z_2i$ then:

$e^{-z}=e^{-(z_1+z_2i)}=e^{-z_1-z_2i}=e^{-z_1}.e^{-z_2i}$

I know the first part is easy to see $e^{-z_1}=\frac{1}{e^{z_1}}$ since $z_1$ is a real number.

However I am not seeing how to proceed with $e^{-z_2i}$.

Question:

Since I have the imaginary unit. How should I treat $e^{-z_2i}$?

Thanks in advance!|

Answer 1

Using $e^{z+w}=e^z\cdot e^w$ you get immediately $$e^{z-z}=e^0 = 1\Rightarrow e^{-z} =\frac{1}{e^z}$$

Answer 2

Hint:$$e^{-z_2i}=(e^{-z_2})^i, z_2\in \mathbb{R}$$

Answer 3

$$\begin{array}{} e^{z_1} e^{z_2}&=\sum_{k_1=0}^\infty\frac{z_1^{k_1}}{k_1!}\sum_{k_2=0}^\infty\frac{z_2^{k_2}}{k_2!}=\sum_{k_1=0}^\infty\sum_{k_2=0}^\infty\frac{z_1^{k_1}z_2^{k_2}}{k_1!k_2!}=\sum_{k=0}^\infty\sum_{k_1=0}^k\frac{z_1^{k_1}z_2^{k-k_1}}{k_1!(k-k_1)!}\\ &=\sum_{k=0}^\infty\frac1{k!}\sum_{k_1=0}^k\frac{k!}{k_1!(k-k_1)!}z_1^{k_1}z_2^{k-k_1}=\sum_{k=0}^\infty\frac{(z_1+z_2)^k}{k!}=e^{z_1+z_2}.\\ \end{array} $$ Subsitute now $z_1=z$, $z_2=-z$.