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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

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

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  • $\begingroup$ Are you implying that the real number($Re$) is equal to zero or you let $Re=0$? $\endgroup$ – behold May 2 at 23:24
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    $\begingroup$ 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

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