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rvc

  • one year ago

Please help. Separate the following complex number into real and imaginary parts.

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  1. rvc
    • one year ago
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    \[\log_{1-i}(1+i) \]

  2. Jhannybean
    • one year ago
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    property of logs?

  3. Michele_Laino
    • one year ago
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    try to change the base please

  4. rvc
    • one year ago
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    \[\frac{\log(1+i)}{log(1-i)}\]

  5. Michele_Laino
    • one year ago
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    here is the formula for base change: \[\Large {\log _{1 - i}}x = \frac{{{{\log }_e}x}}{{{{\log }_e}\left( {1 - i} \right)}}\]

  6. Michele_Laino
    • one year ago
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    if we apply my formula above, when x= 1+i, we will get the requested answer

  7. rvc
    • one year ago
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    ? wait im confused

  8. rvc
    • one year ago
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    what formula?

  9. rvc
    • one year ago
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    please explain

  10. rvc
    • one year ago
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    yes i wrote the same above

  11. hartnn
    • one year ago
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    the thing is its not separated into real and imaginary parts yet

  12. ganeshie8
    • one year ago
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    \[\log_{1-i}(1+i) = \dfrac{\log re^{ix}}{\log re^{-ix}} = \dfrac{\log r + ix}{\log r-ix}\] do the conjugate thingy next

  13. rvc
    • one year ago
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    |dw:1439121116009:dw|

  14. rvc
    • one year ago
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    can i apply that formula?

  15. hartnn
    • one year ago
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    you can apply but then you'll need to do that conjugate thingy which will be complicated with your formula. instead try what ganeshie has suggested.

  16. rvc
    • one year ago
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    after the conversion into polar form what did ganeshie did further

  17. hartnn
    • one year ago
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    know about "rationalizing the denominator" approach? same thing can be done to "real"ize the denominator. multiply numerator and denominator by conjugate of denominator

  18. rvc
    • one year ago
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    im not getting that @hartnn :(

  19. ganeshie8
    • one year ago
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    recall the logarithm properties \[\log ab = \log a+\log b\] and \[\log a^b = b\log a\]

  20. ganeshie8
    • one year ago
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    \[\log_{1-i}(1+i) = \dfrac{\log re^{ix}}{\log re^{-ix}} = \dfrac{\log r+\log e^{ix}}{\log r+\log e^{-ix}} = \dfrac{\log r + ix}{\log r-ix}\] how about now ?

  21. rvc
    • one year ago
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    i did understand you step till\[\rm \frac{ log~re^{ix} }{ log~re^{-ix} }\]

  22. rvc
    • one year ago
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    your*

  23. rvc
    • one year ago
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    got it :) @ganeshie8

  24. ganeshie8
    • one year ago
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    notice that \(\log \color{red}{r}e^{ix}\) is in form \(\log \color{red}{a}b\)

  25. rvc
    • one year ago
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    yep yep after that?

  26. hartnn
    • one year ago
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    multiply numerator and denominator by log r + ix denominator will be of the form (p+iq)(p-iq) = p^2 +q^2

  27. rvc
    • one year ago
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  28. rvc
    • one year ago
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    thanks all :)

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