Complex logarithm rules

  • #1
302
0

Main Question or Discussion Point

Hi.

I know that for real numbers log(z)=-log(1/z)

is this also true in general for complex numbers?
 

Answers and Replies

  • #2
rbj
2,226
8
yup. no promises about the log of 0 or [itex]\infty[/itex].
 
  • #3
2,111
18
But one might need to add [itex]2\pi i[/itex] somewhere sometimes because of some branch choosing issues.
 
  • #4
2,111
18
If you choose to use a branch

[tex]
\log(z) = \log(|z|) + i\textrm{arg}(z),\quad 0\leq \textrm{arg}(z) < 2\pi
[/tex]

then for example

[tex]
\log(-1+i) = \log(\sqrt{2}) + \frac{3\pi i}{4}
[/tex]

and

[tex]
\log(\frac{1}{-1+i}) = \log(-\frac{1}{2}(1+i)) = \log(\frac{1}{\sqrt{2}}) + \frac{5\pi i}{4}.
[/tex]

So you've got

[tex]
\log(-1+i) + \log(\frac{1}{-1+i}) = 2\pi i,
[/tex]

in contradiction with your equation. But if you choose the branch so that

[tex]
-\pi < \textrm{arg}(z) \leq \pi,
[/tex]

then you've got

[tex]
\log(-1+i) + \log(\frac{1}{-1+i}) = 0,
[/tex]

as your equation stated. Even with this choice of branch still, for example,

[tex]
\log(-1) + \log(\frac{1}{-1}) = 2\pi i,
[/tex]

so actually...

Hi.

I know that for real numbers log(z)=-log(1/z)
for positive real numbers! :wink:
 
  • #5
mathman
Science Advisor
7,839
439
But one might need to add [itex]2\pi i[/itex] somewhere sometimes because of some branch choosing issues.

The essentail point is than ln(1)= [itex]2n\pi i[/itex] with n being any integer.
 
  • #6
302
0
Thank you for the answers and examples. I understand it much better now.
 
  • #7
2,111
18
You probably remember the trick where one does something like this:

[tex]
1 = \sqrt{1} = \cdots = -1
[/tex]

with imaginary units. The examples I gave are very similar in nature. Most of the time, a blind use of familiar calculation rules might seem to work, but you never know when something tricky surprises you, if you are not careful.
 

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