Natural logarithm of negative numbers

In summary, the conversation discusses the result that ln(-1/2) = -ln(2) + iπ, and how it can be derived using Euler's formula. The concept of a multivalued logarithm is also mentioned. One of the participants suggests using the complex exponential form of -1/2, z = 1/2e^{iπ(2n+1)}, to arrive at the result. Another participant points out the typo "cis" instead of "cos" and notes that "cis" is shorthand notation for cos θ + i sin θ. Overall, the conversation provides an interesting perspective on the topic.
  • #1
DivGradCurl
372
0
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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  • #2
Use Euler's formula and write "-1/2" in another way involving complex exponentials.Then apply "ln".

Daniel.
 
  • #3
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)
 
  • #4
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...:tongue2:

Daniel.
 
  • #5
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:
  • #6
Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"... :tongue:

Daniel.
 
  • #7
dextercioby said:
Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"... :tongue:

Daniel.

FYI, [tex]cis \theta[/tex] is shorthand notation for [tex]\cos \theta + i \sin \theta[/tex]. It's perfectly valid.
 
  • #8
Hmmmmm...Interesting,even ingenious...

Daniel.
 
  • #9
I haven't seen "cis" in years! (It's more engineer lingo than mathematics.)

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

1. What is the value of the natural logarithm of a negative number?

The natural logarithm of a negative number is undefined. This is because the natural logarithm function is only defined for positive numbers.

2. Can the natural logarithm of a negative number be a complex number?

Yes, the natural logarithm of a negative number can be a complex number. This is because the natural logarithm of a negative number is defined as the complex number whose exponential is the given negative number.

3. Why is the natural logarithm of negative numbers not defined?

The natural logarithm of negative numbers is not defined because the natural logarithm function is only defined for positive numbers. This is due to the mathematical properties of logarithms and the fact that negative numbers do not have real logarithms.

4. Can I use the natural logarithm of negative numbers in real-world applications?

No, the natural logarithm of negative numbers is not applicable in real-world applications. This is because the natural logarithm function is only defined for positive numbers, and real-world applications typically involve positive quantities.

5. Is there a way to approximate the natural logarithm of a negative number?

There is no way to accurately approximate the natural logarithm of a negative number. This is because the natural logarithm function is not defined for negative numbers, so any approximation would be invalid. It is best to avoid using negative numbers with the natural logarithm function.

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