Complex logarithm rules

[daudaudaudau]

Hi.

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

is this also true in general for complex numbers?

[rbj]

yup. no promises about the log of 0 or \infty.

[jostpuur]

But one might need to add 2\pi i somewhere sometimes because of some branch choosing issues.

[jostpuur]

If you choose to use a branch


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


then for example


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


and


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


So you've got


\log(-1+i) + \log(\frac{1}{-1+i}) = 2\pi i,


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


-\pi < \textrm{arg}(z) \leq \pi,


then you've got


\log(-1+i) + \log(\frac{1}{-1+i}) = 0,


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


\log(-1) + \log(\frac{1}{-1}) = 2\pi i,


so actually...

Hi.

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



for positive real numbers! :wink:

[mathman]

But one might need to add 2\pi i somewhere sometimes because of some branch choosing issues.


The essentail point is than ln(1)= 2n\pi i with n being any integer.

[daudaudaudau]

Thank you for the answers and examples. I understand it much better now.

[jostpuur]

You probably remember the trick where one does something like this:


1 = \sqrt{1} = \cdots = -1


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.