Are complex logarithm rules the same as real number 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

<br /> \log(z) = \log(|z|) + i\textrm{arg}(z),\quad 0\leq \textrm{arg}(z) &lt; 2\pi<br />

then for example

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

and

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

So you've got

<br /> \log(-1+i) + \log(\frac{1}{-1+i}) = 2\pi i,<br />

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

<br /> -\pi &lt; \textrm{arg}(z) \leq \pi,<br />

then you've got

<br /> \log(-1+i) + \log(\frac{1}{-1+i}) = 0,<br />

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

<br /> \log(-1) + \log(\frac{1}{-1}) = 2\pi i,<br />

so actually...

Hi.

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

for positive real numbers!

 

[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:

<br /> 1 = \sqrt{1} = \cdots = -1<br />

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.