ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: brightsky on January 06, 2010, 05:55:24 pm

Title: Quick Question
Post by: brightsky on January 06, 2010, 05:55:24 pm: How do you solve $\log_e(-5)$ considering complex number solutions?
Title: Re: Quick Question
Post by: jimmy999 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 :)
Title: Re: Quick Question
Post by: /0 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 $\theta \in (-\pi, \pi]$
Title: Re: Quick Question
Post by: brightsky 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
Title: Re: Quick Question
Post by: /0 on January 06, 2010, 08:39:52 pm: Oh, i forgot to mention:

$re^{i\theta}=r(\cos{\theta}+i\sin{\theta})$
Title: Re: Quick Question
Post by: TrueTears on January 06, 2010, 10:15:14 pm: Same as cis.