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?

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- Dec 18th 2007, 03:26 PMFailbaitNatural 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? - Dec 18th 2007, 04:05 PMThePerfectHacker
- Dec 18th 2007, 04:24 PMFailbait
ln(-.3675) = $\displaystyle -1.001+\pi i$

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

Do those answers look about right?

Thanks! - Dec 19th 2007, 01:35 AMmr fantastic
- Dec 19th 2007, 03:20 AMwingless
$\displaystyle e^{i\theta} = cos \theta + i sin \theta$

$\displaystyle e^{i\pi} = cos \pi + i sin \pi$

$\displaystyle e^{i\pi} = -1$

$\displaystyle ln(e^{i\pi}) = ln -1$

$\displaystyle i\pi = ln -1$

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. - Dec 19th 2007, 07:33 AMThePerfectHacker
- Dec 19th 2007, 12:42 PMwingless
- Dec 19th 2007, 02:05 PMThePerfectHacker
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 $\displaystyle z_1=z_2=-1$. :eek:

Complex logarithm.