MHB Natural log of a complex number.

Drain Brain
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Evaluate the following logarithms, expressing the answers in rectangular form

a. $\ln1$, $Ln1$
b. $\ln(3-j4)$, $Ln(3-j4)$

I know that the log of a complex number z is given as

$\ln z=\ln|z|+argz$

but I still don't know how to use this fact to solve the problems above. I'm having a hard time understanding the material that I read about this. please enlighten me.
 
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Drain Brain said:
Evaluate the following logarithms, expressing the answers in rectangular form

a. $\ln1$, $Ln1$
b. $\ln(3-j4)$, $Ln(3-j4)$

I know that the log of a complex number z is given as

$\ln z=\ln|z|+argz$

but I still don't know how to use this fact to solve the problems above. I'm having a hard time understanding the material that I read about this. please enlighten me.

Just a Question: what is the difference in the notations $\ln\ a$ and $\text{Ln}\ a$, being a a complex variable? ...

Kind regards

$\chi$ $\sigma$

- - - Updated - - -

Drain Brain said:
Evaluate the following logarithms, expressing the answers in rectangular form$\ln z=\ln|z|+arg z$

What I remember is that, setting $\displaystyle j = \sqrt{-1}$, is $\displaystyle \ln z = \ln |z| + j\ \text{arg}\ z$...

Kind regards

$\chi$ $\sigma$
 
Drain Brain said:
Evaluate the following logarithms, expressing the answers in rectangular form

a. $\ln1$, $Ln1$
b. $\ln(3-j4)$, $Ln(3-j4)$

I know that the log of a complex number z is given as

$\ln z=\ln|z|+argz$

but I still don't know how to use this fact to solve the problems above. I'm having a hard time understanding the material that I read about this. please enlighten me.

Surely you can do ln(1) with your eyes closed.

As for the rest, can't you evaluate |3 - 4j| and arg(3 - 4j) ?
 
chisigma said:
Just a Question: what is the difference in the notations $\ln\ a$ and $\text{Ln}\ a$, being a a complex variable? ...

Kind regards

$\chi$ $\sigma$

- - - Updated - - -
What I remember is that, setting $\displaystyle j = \sqrt{-1}$, is $\displaystyle \ln z = \ln |z| + j\ \text{arg}\ z$...

Kind regards

$\chi$ $\sigma$

the $Ln z$ is the principal value of $\ln z$
 
Drain Brain said:
the $Ln z$ is the principal value of $\ln z$

That means that is $\displaystyle \ln z = \text{Ln}\ z + 2\ k\ \pi\ i$... so that if you know you have the other one automatically... all right?...

Kind regards

$\chi$ $\sigma$
 

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