anonymous
  • anonymous
Evaluate Log e (2+J5) giving the answer in the form a+jb .
Mathematics
chestercat
  • chestercat
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dumbcow
  • dumbcow
1.68 + j1.19
anonymous
  • anonymous
you looking for this?\[ \ln(2+J5)\]
dumbcow
  • dumbcow
yeah thats what i assumed

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anonymous
  • anonymous
so how did you get the answer..
dumbcow
  • dumbcow
haha calculator :)
anonymous
  • anonymous
ln2+...
anonymous
  • anonymous
how did you use calculater..
anonymous
  • anonymous
you know that right.. \[\ln(a+b)\neq lna+lnb\]
dumbcow
  • dumbcow
well you would have to have a graphing calculator (TI 83 or better) ln(2+5i) , there is symbol for an imaginary number
dumbcow
  • dumbcow
yes i am aware of that
anonymous
  • anonymous
but it is not i, it is j
anonymous
  • anonymous
never mind then..
dumbcow
  • dumbcow
same thing, different notation for a complex number which has real part and imaginary part
anonymous
  • anonymous
I've never seen a+jb http://en.wikipedia.org/wiki/Complex_number
anonymous
  • anonymous
i assume this is a complex number
anonymous
  • anonymous
usually written as \[\ln(a+bi)\] or more usually \[\log(a+bi)\]
anonymous
  • anonymous
find it via \[\log(a+bi)=\ln(|a+bi|)+i\theta\] where \[\tan(\theta)=\frac{b}{a}\] and \[|a+bi|=\sqrt{a^2+b^2}\]
anonymous
  • anonymous
I like this one
anonymous
  • anonymous
can we say that then..\[|a+bi|=\sqrt{x^2+b^2}=\sqrt{29}\] \[\tan^{-1}( 5/2)=68=\theta\] \[\log(2+5i)=\ln|2+5i|+i \theta\] =\[\ln \sqrt{29}+i68=1.68+68i\]
anonymous
  • anonymous
actually thisone is better\[\log(a+bi)=\log(|a+bi|)+i\theta\quad or\quad \ln(a+bi)=\ln(|a+bi|)+i\theta\]
anonymous
  • anonymous
68/180=R/pi \[68 \quad degree \approx \frac25\pi\] \[\log_{e} (2+5i)=\ln(2+5i)=1.68+\frac25\pi+2k \pi i\]

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