How do you prove a natural logarithm with complex numbers equal to a natural logarithm with an absolute value?

Question

The author stated that if $ln$ $z=a+ib$ then $ln |z|=a$
Can someone show me a proof of this? I have been looking and can not find one to see if its true

What I do see is that if $z=a+ib$ then the $a$ and $b$ are real and imaginary parts respectively which can be written as $Re(z)=a$ and $Im(z)=b$, note: $Im$ means imaginary.

I think the author trying to say that a real and an imaginary number of a natural logarithm equals a logarithm with a real number. I can not show this

Answer 1

Asserting that $a+bi$ is a logarithm of $z$ is the same thing as asserting that $e^{a+bi}=z$. But then\begin{align}\lvert z\rvert&=\lvert e^{a+bi}\rvert\\&=\lvert e^a\times e^{bi}\rvert\\&=\lvert e^a\rvert\times\lvert e^{bi}\rvert\\&=e^a,\end{align}since $\lvert e^{bi}\rvert=e^{\operatorname{Re}(bi)}=e^0=1$. But then $\ln\lvert z\rvert=a$.