Complex logarithm rules

1. Nov 2, 2008

daudaudaudau

Hi.

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

is this also true in general for complex numbers?

2. Nov 2, 2008

rbj

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

3. Nov 2, 2008

jostpuur

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

4. Nov 2, 2008

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...

for positive real numbers!

5. Nov 2, 2008

mathman

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

6. Nov 2, 2008

daudaudaudau

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

7. Nov 2, 2008

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.