Why $Log(e^{z})=z$

Question

$Log(z)=log(\left | z \right |)+iArg(z)$ where $\left \{ z \in \mathbb{C}:-\pi <Imz\leq \pi \right \}$

Now $Log(e^{z})=log(e^x)+iArg(e^{x+iy})=x+iArg(e^{x+iy})$
my question is why $Arg(e^{x+iy})=y$?
So it can be $Log(e^{z})=z$

Answer 1

Notice that : $$ \mathrm{e}^{x+\mathrm{i}y}=r\,\mathrm{e}^{\mathrm{i}\theta} $$

were $ r $ here is $ \mathrm{e}^{x}>0 $, and $ \theta $ is none other than $ y\in\mathbb{R} $.

Therefore : $ \arg\left(\mathrm{e}^{x+\mathrm{i}y}\right)\equiv\theta \equiv y\ \left(\mathrm{mod}\ 2\pi\right) $.