Natural logarithm of negative numbers

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Homework Help Overview

The discussion revolves around the natural logarithm of negative numbers, specifically the expression involving the logarithm of \(-\frac{1}{2}\) and its relationship to complex numbers. Participants explore the implications of using Euler's formula and the multivalued nature of the logarithm function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of negative numbers into complex exponential form and the application of logarithmic properties. Questions arise regarding the validity of certain logarithmic expressions and the implications of multivalued functions.

Discussion Status

The conversation is active, with participants providing insights into the use of Euler's formula and the nature of logarithms in complex analysis. Some guidance has been offered regarding the general formula for logarithms of complex numbers, though multiple interpretations are being explored.

Contextual Notes

There is mention of the notation "cis" and its relevance in the context of complex numbers, highlighting a potential gap in familiarity among participants. The discussion also touches on the informal nature of some terminology used in engineering versus mathematics.

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

Daniel.
 
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)
 
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...:-p

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

Daniel.
 
dextercioby said:
Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"... :-p

Daniel.

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

Daniel.
 
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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