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

  • $\begingroup$ Are you implying that the real number($Re$) is equal to zero or you let $Re=0$? $\endgroup$ – behold May 2 at 23:24
  • 1
    $\begingroup$ I am asserting that, if $b\in\mathbb R$, then $\operatorname{Re}(bi)=0$. $\endgroup$ – José Carlos Santos May 2 at 23:26
  • $\begingroup$ Oh I got you. I was wondering about that too. thanks $\endgroup$ – behold May 2 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.