natural logarithm of negative numbers

DivGradCurl

I'm puzzled by....

[tex]\ln \left( -\frac{1}{2} \right) = - \ln \left( 2 \right) + i \pi [/tex]

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

I know that

[tex]\ln \left( \frac{1}{2} \right) = - \ln \left( 2 \right)[/tex].

Thank you so much

dextercioby

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

Daniel.

Hurkyl

Technically, that's not true. ln is a multivalued function. In particular,

[tex]
\ln -\frac{1}{2} = \ln \frac{1}{2} + i \pi (1 + 2n) \quad (n \in \mathbb{Z})
[/tex]

Here is the general formula, which can be derived as Daniel suggests.

[tex]
\ln z = \ln |z| + i \arg z
[/tex]

(where the ln on the RHS is the ordinary real-valued logarithm)

dextercioby

Hurkyl,i was suggesting:

[tex] -\frac{1}{2}=-1 \cdot \frac{1}{2} [/tex]

And of course:

[tex] -1=e^{i\pi (2n+1)} \ (n\in \mathbb{Z}) [/tex]

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

Daniel.

DivGradCurl

That makes sense! So, if

[tex]z=-\frac{1}{2}=\frac{1}{2}\mbox{ cis }\pi (2n+1) = \frac{1}{2} e^{i\pi (2n+1)}[/tex]

Then

[tex]\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}) [/tex]

Thanks

dextercioby

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

Daniel.

Curious3141

FYI, [tex]cis \theta[/tex] is shorthand notation for [tex]\cos \theta + i \sin \theta[/tex]. It's perfectly valid.

dextercioby

Hmmmmm...Interesting,even ingenious...

Daniel.

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.)