## natural logarithm of negative numbers

I'm puzzled by....

$$\ln \left( -\frac{1}{2} \right) = - \ln \left( 2 \right) + i \pi$$

Why is this true? How can I possibly get this result?

I know that

$$\ln \left( \frac{1}{2} \right) = - \ln \left( 2 \right)$$.

Thank you so much
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Use Euler's formula and write "-1/2" in another way involving complex exponentials.Then apply "ln". Daniel.
 Recognitions: Gold Member Science Advisor Staff Emeritus Technically, that's not true. ln is a multivalued function. In particular, $$\ln -\frac{1}{2} = \ln \frac{1}{2} + i \pi (1 + 2n) \quad (n \in \mathbb{Z})$$ Here is the general formula, which can be derived as Daniel suggests. $$\ln z = \ln |z| + i \arg z$$ (where the ln on the RHS is the ordinary real-valued logarithm)

Blog Entries: 9
Recognitions:
Homework Help

## natural logarithm of negative numbers

Hurkyl,i was suggesting:

$$-\frac{1}{2}=-1 \cdot \frac{1}{2}$$

And of course:

$$-1=e^{i\pi (2n+1)} \ (n\in \mathbb{Z})$$

No need to know any definition,just Euler's formula...

Daniel.
 That makes sense! So, if $$z=-\frac{1}{2}=\frac{1}{2}\mbox{ cis }\pi (2n+1) = \frac{1}{2} e^{i\pi (2n+1)}$$ Then $$\ln z = \ln \left[ \frac{1}{2} e^{i\pi (2n+1)} \right] = -\ln 2\mbox{ } +\mbox{ } i\pi (2n+1) \qquad (n\in \mathbb{Z})$$ Thanks
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"... Daniel.

Recognitions:
Homework Help
 Quote by dextercioby Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"... Daniel.
FYI, $$cis \theta$$ is shorthand notation for $$\cos \theta + i \sin \theta$$. It's perfectly valid.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Hmmmmm...Interesting,even ingenious... Daniel.
 Recognitions: Gold Member Science Advisor Staff Emeritus I haven't seen "cis" in years! (It's more engineer lingo than mathematics.) (And both dextercioby's posts were clearly tongue in cheek.)