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