# Homework Help: Natural logarithm of negative numbers

1. Feb 17, 2005

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

2. Feb 17, 2005

### dextercioby

Use Euler's formula and write "-1/2" in another way involving complex exponentials.Then apply "ln".

Daniel.

3. Feb 17, 2005

### Hurkyl

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)

4. Feb 17, 2005

### dextercioby

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...:tongue2:

Daniel.

5. Feb 17, 2005

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

Last edited: Feb 17, 2005
6. Feb 18, 2005

### dextercioby

Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"... :tongue:

Daniel.

7. Feb 18, 2005

### Curious3141

FYI, $$cis \theta$$ is shorthand notation for $$\cos \theta + i \sin \theta$$. It's perfectly valid.

8. Feb 18, 2005

### dextercioby

Hmmmmm...Interesting,even ingenious...

Daniel.

9. Feb 18, 2005

### HallsofIvy

I haven't seen "cis" in years! (It's more engineer lingo than mathematics.)

(And both dextercioby's posts were clearly tongue in cheek.)