# Thread: Natural log of a negative number

1. ## 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?

2. 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 $\displaystyle z\not = 0$ then $\displaystyle \log z = \log |z| + i\arg (z)$

Can you do that?

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

Thanks!

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

Thanks!
Hmmmm, let's see now ..... does |-0.3675| = -1.001? Does |-13.4| = 2.595?

5. $\displaystyle e^{i\theta} = cos \theta + i sin \theta$
$\displaystyle e^{i\pi} = cos \pi + i sin \pi$
$\displaystyle e^{i\pi} = -1$
$\displaystyle ln(e^{i\pi}) = ln -1$
$\displaystyle i\pi = ln -1$

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

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

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

Yes, they're correct.

6. Originally Posted by wingless
$\displaystyle ln(e^{i\pi}) = ln -1$
$\displaystyle i\pi = ln -1$
You are not allowed to do that. The rule $\displaystyle \log e^z = z$ for $\displaystyle z\in \mathbb{C}^{\text{x}}$ does not work in general.
However, $\displaystyle e^{\log z} = z$ does work in general.

7. Originally Posted by ThePerfectHacker
You are not allowed to do that. The rule $\displaystyle \log e^z = z$ for $\displaystyle z\in \mathbb{C}^{\text{x}}$ 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 $\displaystyle ln -1 = i \pi$.

8. Originally Posted by wingless
Ok, the formula is still true since $\displaystyle ln -1 = i \pi$.
Yes! Sometimes the formula will work sometimes it does not.
Another thing that breaks down is $\displaystyle \log (z_1z_2) = \log z_1 + \log z_2$.
Take for example $\displaystyle z_1=z_2=-1$.
Complex logarithm.