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Callisto

  • 2 years ago

Complex number. For the exponential form of a complex number, \(a + bi = e^{\log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}\), why is it using common log, but not natural log or log with other base?

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  1. experimentX
    • 2 years ago
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    no quite ... !!

  2. Callisto
    • 2 years ago
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    ...... Yes?!

  3. experimentX
    • 2 years ago
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    |dw:1345042462489:dw|

  4. experimentX
    • 2 years ago
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    |dw:1345042511668:dw|

  5. experimentX
    • 2 years ago
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    |dw:1345042599891:dw|

  6. experimentX
    • 2 years ago
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    |dw:1345042661317:dw|

  7. experimentX
    • 2 years ago
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    sometimes ... we need r and e^itheta as one ... |dw:1345042767656:dw|

  8. Callisto
    • 2 years ago
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    |dw:1345042836751:dw|

  9. experimentX
    • 2 years ago
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    |dw:1345042823054:dw| yep ... my bad :))

  10. Callisto
    • 2 years ago
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    So, is it ln or log?

  11. experimentX
    • 2 years ago
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    ln .. since in usually draw ... ln seems like in .. and one user mistook for lnt for int

  12. experimentX
    • 2 years ago
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    using this ... try to find the value of i^i

  13. Callisto
    • 2 years ago
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    Your show time again :|

  14. experimentX
    • 2 years ago
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    |dw:1345043073075:dw|

  15. Callisto
    • 2 years ago
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    Not quite understand..

  16. experimentX
    • 2 years ago
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    |dw:1345043197633:dw|

  17. Callisto
    • 2 years ago
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    \[a + bi = e^{\ln \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}\]Put a = 0 and b = 1 into the formula. \[i = 0 + 1i \]\[= e^{\ln \sqrt{0^2 + 1^2} + i (tan^{-1}\frac{1}{0})}\]\[=e^{\ln (1) + i (tan^{-1}(undefined))}\]\[=e^{i (tan^{-1}(undefined))}\]It doesn't seem right at all!

  18. experimentX
    • 2 years ago
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    that would be pi/2

  19. Callisto
    • 2 years ago
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    I know... but that would be \[=e^{\frac{i\pi}{2}}\]Something goes wrong here?!

  20. experimentX
    • 2 years ago
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    it's all right ... now complete the problem.

  21. Callisto
    • 2 years ago
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    May I know.. how?

  22. experimentX
    • 2 years ago
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    |dw:1345043811417:dw|

  23. Callisto
    • 2 years ago
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    Oh! \[i^i = (e^{\frac{i\pi}{2}})^{i}=(e^{\frac{i^2\pi}{2}})=e^{\frac{-\pi}{2}}\] The problem was that I didn't know what I was doing!

  24. experimentX
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=i^i+%3D+e^%28-pi%2F2%29 you are right!!

  25. Callisto
    • 2 years ago
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    Thanks a ton!!! :)

  26. experimentX
    • 2 years ago
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    yw!! you can simplify things of the form log(a+ib)

  27. Callisto
    • 2 years ago
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    That's another thing?

  28. experimentX
    • 2 years ago
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    no not really ... http://www.wolframalpha.com/input/?i=log%281+%2B+isqrt%283%29%29

  29. Callisto
    • 2 years ago
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    *lost* If \(a + bi = e^{\log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}\) Then, \(\ln (a + bi) = \ln (e^{\log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})})\) \(\ln (a + bi) = \log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})\) ?! Looks strange enough to me :|

  30. experimentX
    • 2 years ago
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    yep ... that's right!!

  31. Callisto
    • 2 years ago
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    Thanks again!

  32. experimentX
    • 2 years ago
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    np :) you might find this interesting http://www.wolframalpha.com/input/?i=sin%283+%2B+i4%29

  33. experimentX
    • 2 years ago
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    basically ... |dw:1345045135680:dw| using this identity ... you can prove any trigonometric identity you did back in school. though redoing them again ... is crazy.

  34. Callisto
    • 2 years ago
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    Never mind... I guess I'll spend some time re-doing them again.. To see if I really understand! Thanks!

  35. experimentX
    • 2 years ago
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    yw!! best of luck!!

  36. mahmit2012
    • 2 years ago
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    |dw:1348318962769:dw|

  37. mahmit2012
    • 2 years ago
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    |dw:1348319040083:dw|

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