Why Is $\operatorname{Ln}(z)$ The Inverse Of $e^z$

Question

$$\operatorname{Ln}(z)=\ln|z|+i\operatorname{Arg}(z)$$

so taking it as the power of $e$ we get

$$e^{\operatorname{Ln}(z)}=e^{\ln|z|+i\operatorname{Arg}(z)}=e^{\ln|z|}\cdot e^{i\operatorname{Arg}(z)}=|z|\cdot \operatorname{cis}(\operatorname{Arg}(z))$$

How do we arrive to $z$ from there?

Answer 1

$e^{Ln(z)}=e^{ln|z|+iArg(z)}=e^{ln|z|}e^{iArg(z)}=|z|cis(Arg(z))=^{(*)}z$

(*) - Any complex number $z=a+ib$ can be represented as $z=r*(cos(\theta)+isin(\theta))$ where $r=\sqrt{a^2+b^2} , \theta=arctan(\frac b a)$.

This representation is called "Polar Representation", and can be derived by looking at the complex plane as a coordinate system where the vertical axis represents the $i$ component of $z$ and the horizonal axis represents the real component of z.

Answer 2

$$|z|\text{ cis}\arg z=\sqrt{x^2+y^2}\text{ cis}\left(\arctan\frac yx\right).$$

Then from

$$\cos\theta=\frac1{\sqrt{\tan^2\theta+1}},\sin\theta=\frac{\tan\theta}{\sqrt{\tan^2\theta+1}}$$ we draw

$$\text{cis}\left(\arctan\frac yx\right)=\frac{1+i\dfrac yx}{\sqrt{1+\dfrac{y^2}{x^2}}}$$

and finally

$$|z|\text{ cis}\arg z=x+iy.$$