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anonymous

  • 4 years ago

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

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  1. dumbcow
    • 4 years ago
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    1.68 + j1.19

  2. anonymous
    • 4 years ago
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    you looking for this?\[ \ln(2+J5)\]

  3. dumbcow
    • 4 years ago
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    yeah thats what i assumed

  4. anonymous
    • 4 years ago
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    so how did you get the answer..

  5. dumbcow
    • 4 years ago
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    haha calculator :)

  6. anonymous
    • 4 years ago
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    ln2+...

  7. anonymous
    • 4 years ago
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    how did you use calculater..

  8. anonymous
    • 4 years ago
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    you know that right.. \[\ln(a+b)\neq lna+lnb\]

  9. dumbcow
    • 4 years ago
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    well you would have to have a graphing calculator (TI 83 or better) ln(2+5i) , there is symbol for an imaginary number

  10. dumbcow
    • 4 years ago
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    yes i am aware of that

  11. anonymous
    • 4 years ago
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    but it is not i, it is j

  12. anonymous
    • 4 years ago
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    never mind then..

  13. dumbcow
    • 4 years ago
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    same thing, different notation for a complex number which has real part and imaginary part

  14. anonymous
    • 4 years ago
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    I've never seen a+jb http://en.wikipedia.org/wiki/Complex_number

  15. anonymous
    • 4 years ago
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    i assume this is a complex number

  16. anonymous
    • 4 years ago
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    usually written as \[\ln(a+bi)\] or more usually \[\log(a+bi)\]

  17. anonymous
    • 4 years ago
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    find it via \[\log(a+bi)=\ln(|a+bi|)+i\theta\] where \[\tan(\theta)=\frac{b}{a}\] and \[|a+bi|=\sqrt{a^2+b^2}\]

  18. anonymous
    • 4 years ago
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    I like this one

  19. anonymous
    • 4 years ago
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    can we say that then..\[|a+bi|=\sqrt{x^2+b^2}=\sqrt{29}\] \[\tan^{-1}( 5/2)=68=\theta\] \[\log(2+5i)=\ln|2+5i|+i \theta\] =\[\ln \sqrt{29}+i68=1.68+68i\]

  20. anonymous
    • 4 years ago
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    actually thisone is better\[\log(a+bi)=\log(|a+bi|)+i\theta\quad or\quad \ln(a+bi)=\ln(|a+bi|)+i\theta\]

  21. anonymous
    • 4 years ago
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    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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