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

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

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 $$\ln\lvert z\rvert=a$$.

• Are you implying that the real number($Re$) is equal to zero or you let $Re=0$? – behold May 2 at 23:24
• I am asserting that, if $b\in\mathbb R$, then $\operatorname{Re}(bi)=0$. – José Carlos Santos May 2 at 23:26
• Oh I got you. I was wondering about that too. thanks – behold May 2 at 23:27