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rvc

  • one year ago

separate into real and imaginary parts

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  1. rvc
    • one year ago
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    \[\sin^{-1} (e^{i\theta})\]

  2. rvc
    • one year ago
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    \[e^{i\theta}= \sin(x+iy)\]

  3. rvc
    • one year ago
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    \[\cos\theta+isin\theta=sinxcos(iy)+cosxsin(iy)\] \[= sinxcoshy+icosxsinhy\]

  4. rvc
    • one year ago
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    \[\cos\theta=sinxcoshy~;~\sin\theta=cosxsinhy\]

  5. rvc
    • one year ago
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    correct?

  6. ribhu
    • one year ago
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    hey

  7. rvc
    • one year ago
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    please help nn :)

  8. ribhu
    • one year ago
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    can i help? if u want?

  9. rvc
    • one year ago
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    yep

  10. rvc
    • one year ago
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    please ans

  11. ribhu
    • one year ago
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    remember me?

  12. dinamix
    • one year ago
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    i think \[\sin^{-1} (\sin(x+iy)) = x+iy\]

  13. dinamix
    • one year ago
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    @rvc

  14. dinamix
    • one year ago
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    did u see my solution

  15. rvc
    • one year ago
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    no its sin inverse e^itheta

  16. rvc
    • one year ago
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    @ribhu if u want to help then i dont mind dont ask questions like that

  17. dinamix
    • one year ago
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    u didnt mean arcsin or what

  18. rvc
    • one year ago
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    \[Let~\sin^{-1} e^{i\theta}=x+iy\]

  19. dinamix
    • one year ago
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    cuz i know sin^-1 = arcsin

  20. IrishBoy123
    • one year ago
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    try starting with \[sin z = \frac{e^z - e^{-z}}{2i}\] and note that \[w = arcsin \ e^{i \theta} \implies sin(w) = \ e^{i \theta} =\frac{e^w - e^{-w}}{2i} \] ugly but potentially do-able

  21. rvc
    • one year ago
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    hmm what if i go with my steps

  22. rvc
    • one year ago
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    @hartnn pls help

  23. IrishBoy123
    • one year ago
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    you steps are technically correct, i believe. where are you taking it from here?

  24. rvc
    • one year ago
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    i need the value of x and y

  25. IrishBoy123
    • one year ago
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    well i found a reliable formula \[arcsin z = -i \ ln (iz \pm \sqrt{1-z^2})\] and the answer i get from the method i propose actually looks very similar i do not see where you go next in you approach though i see no problems with it

  26. IrishBoy123
    • one year ago
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    following on: \(2i e^{i \theta } = e^w - e^{-w}\) \(i = e^{\pi/2}\) \(u = e^w\) \(u - 2 e^{i(\theta + \pi/2)} - \frac{1}{u} = 0\) \(u^2 - u.2 e^{i(\theta + \pi/2)} - 1 = 0\) and apply quadratic formula => gets you that really ugly formula i posted above

  27. IrishBoy123
    • one year ago
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    the switch to \(e^{i(\theta + \pi /2 )}\) is OTT and only complicates matters. you can just stuff it all straight into the quadratic mea culpa :-(

  28. rvc
    • one year ago
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    @hartnn @baru @IrishBoy123 can we get back to this question please..

  29. baru
    • one year ago
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    i dont even understand how to interpret that, sin( complex number) is a thing ?

  30. baru
    • one year ago
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    i think these are @ParthKohli 's hunting grounds

  31. rvc
    • one year ago
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    @Kainui

  32. rvc
    • one year ago
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    @hartnn please help whenever u r online...

  33. Kainui
    • one year ago
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    $$z= \frac{z+z^*}{2} + i \frac{z-z^*}{2i} = \Re(z)+i \Im(z)$$ That's probably the most important formula you can have, try it out on anything simple to test it.

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