# Natural log of a negative number

• December 18th 2007, 03:26 PM
Failbait
Natural log of a negative number
I don't understand how to grasp such a concept. Obviously the answer is in a+bi form, but I have no idea how to get there. :(

ln(-.3675)
ln(-13.4)

Here are a few problems from my book, can anyone guide me through?
• December 18th 2007, 04:05 PM
ThePerfectHacker
Quote:

Originally Posted by Failbait
I don't understand how to grasp such a concept. Obviously the answer is in a+bi form, but I have no idea how to get there. :(

ln(-.3675)
ln(-13.4)

Here are a few problems from my book, can anyone guide me through?

If $z\not = 0$ then $\log z = \log |z| + i\arg (z)$

Can you do that?
• December 18th 2007, 04:24 PM
Failbait
ln(-.3675) = $-1.001+\pi i$
ln(-13.4) = $2.595+\pi i$

Thanks!
• December 19th 2007, 01:35 AM
mr fantastic
Quote:

Originally Posted by Failbait
ln(-.3675) = $-1.001+\pi i$
ln(-13.4) = $2.595+\pi i$

Thanks!

Hmmmm, let's see now ..... does |-0.3675| = -1.001? Does |-13.4| = 2.595?
• December 19th 2007, 03:20 AM
wingless
$e^{i\theta} = cos \theta + i sin \theta$
$e^{i\pi} = cos \pi + i sin \pi$
$e^{i\pi} = -1$
$ln(e^{i\pi}) = ln -1$
$i\pi = ln -1$

For $x \in R^+$
$ln -x = ln -1 + ln x = ln x + i\pi$

This is the same thing as ThePerfectHacker's formula. Remember that mine only works for real numbers.

ln(-.3675) = $-1.001+\pi i$
ln(-13.4) = $2.595+\pi i$

Yes, they're correct.
• December 19th 2007, 07:33 AM
ThePerfectHacker
Quote:

Originally Posted by wingless
$ln(e^{i\pi}) = ln -1$
$i\pi = ln -1$

You are not allowed to do that. The rule $\log e^z = z$ for $z\in \mathbb{C}^{\text{x}}$ does not work in general.
However, $e^{\log z} = z$ does work in general.
• December 19th 2007, 12:42 PM
wingless
Quote:

Originally Posted by ThePerfectHacker
You are not allowed to do that. The rule $\log e^z = z$ for $z\in \mathbb{C}^{\text{x}}$ does not work in general.

Wow, I didn't know that. Sorry for that mistake, thanks for correcting me. Ok, the formula is still true since $ln -1 = i \pi$.
• December 19th 2007, 02:05 PM
ThePerfectHacker
Quote:

Originally Posted by wingless
Ok, the formula is still true since $ln -1 = i \pi$.

Yes! Sometimes the formula will work sometimes it does not.
Another thing that breaks down is $\log (z_1z_2) = \log z_1 + \log z_2$.
Take for example $z_1=z_2=-1$. :eek:
Complex logarithm.