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anonymous
 4 years ago
Evaluate Log e (2+J5) giving the answer in the form a+jb .
anonymous
 4 years ago
Evaluate Log e (2+J5) giving the answer in the form a+jb .

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you looking for this?\[ \ln(2+J5)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah thats what i assumed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how did you get the answer..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did you use calculater..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you know that right.. \[\ln(a+b)\neq lna+lnb\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well you would have to have a graphing calculator (TI 83 or better) ln(2+5i) , there is symbol for an imaginary number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i am aware of that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but it is not i, it is j

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0same thing, different notation for a complex number which has real part and imaginary part

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I've never seen a+jb http://en.wikipedia.org/wiki/Complex_number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i assume this is a complex number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0usually written as \[\ln(a+bi)\] or more usually \[\log(a+bi)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0find 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
 4 years ago
Best ResponseYou've already chosen the best response.0can we say that then..\[a+bi=\sqrt{x^2+b^2}=\sqrt{29}\] \[\tan^{1}( 5/2)=68=\theta\] \[\log(2+5i)=\ln2+5i+i \theta\] =\[\ln \sqrt{29}+i68=1.68+68i\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually thisone is better\[\log(a+bi)=\log(a+bi)+i\theta\quad or\quad \ln(a+bi)=\ln(a+bi)+i\theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.068/180=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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