Author Topic: Quick Question (Read 852 times) Tweet Share

brightsky · « **on:** January 06, 2010, 05:55:24 pm »

How do you solve $\log_e(-5)$ considering complex number solutions?

jimmy999 · « **Reply #1 on:** January 06, 2010, 06:27:21 pm »

Ok so using the identity $e^{i\pi} = -1$
We can rearrange this to get $i\pi = \log_e({-1})$

Now $log_e(-5) = log_e(5) + log_e(-1)$
Therefore $log_e(-5) = log_e(5) + i\pi$

I love how I understand that now

/0 · « **Reply #2 on:** January 06, 2010, 06:30:37 pm »

It's not part of the specialist maths course.

In general though, if $z = re^{i \theta}$ is a complex number, then

$\log_e(z) = \log_e{(re^{i\theta})}=\log_e{r}+i\theta$

The principle value of the logarithm is given by $\theta \in (-\pi, \pi]$

brightsky · « **Reply #3 on:** January 06, 2010, 07:01:55 pm »

Oh right, thanks guys! Hehe, I wish I could understand what $z = re^{i\theta}$ meant \0. :p

/0 · « **Reply #4 on:** January 06, 2010, 08:39:52 pm »

Oh, i forgot to mention:

$re^{i\theta}=r(\cos{\theta}+i\sin{\theta})$

TrueTears · « **Reply #5 on:** January 06, 2010, 10:15:14 pm »

Same as cis.

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