Show that $e^{\ln(z)}$ always equals $z$, but $\ln(e^z)$ does not always equal $z$.

Question

Using, Euler's identity, we get:
$z = r.e^{i\theta}$
Then, do we need to use the fact: $\theta = \theta + 2n \pi, n \in \mathbb N$ ?
I am unable to get to the proof..

Answer 1

The equality in the first part follows by definition because whatever $z$ is, we have by definition that $e^{\log z}=z,$ even when $\log z$ has multiple values, since they only differ by some multiple of $2π$ when written out in trigonometric form.

However, for exactly the same reason, the logarithm being multivalued does not return a single value for any argument, and in particular for the well-defined argument $e^z,$ being complex. Thus, there is no unique value of $\log (e^z).$

Answer 2

We define the principal natural logarithm $\ln z$ of any $z\in\Bbb C\setminus\{0\}$ as one solution of $e^w=z$, so by definition $e^{\ln z}=z$. But $\ln(e^z)$ must have period $2\pi i$, so can't be the identity function.