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natural logarithm of negative numbers |
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| Feb17-05, 03:46 PM | #1 |
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natural logarithm of negative numbers
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 |
| Feb17-05, 03:57 PM | #2 |
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Use Euler's formula and write "-1/2" in another way involving complex exponentials.Then apply "ln".
Daniel. |
| Feb17-05, 04:04 PM | #3 |
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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) |
| Feb17-05, 04:10 PM | #4 |
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natural logarithm of negative numbers
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. |
| Feb17-05, 05:16 PM | #5 |
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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 |
| Feb18-05, 12:41 AM | #6 |
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Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"...
![]() Daniel. |
| Feb18-05, 03:28 AM | #7 |
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Recognitions:
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| Feb18-05, 04:25 AM | #8 |
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Hmmmmm...Interesting,even ingenious...
Daniel. |
| Feb18-05, 05:48 AM | #9 |
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I haven't seen "cis" in years! (It's more engineer lingo than mathematics.)
(And both dextercioby's posts were clearly tongue in cheek.) |
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