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Loser66

  • one year ago

Find z such that tanz = i/2 Please, help

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  1. Loser66
    • one year ago
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    My attempt: \(tan z = \dfrac{sin z}{cos z}=\dfrac{e^{iz}-e^{-iz}}{2i} *\dfrac{2}{e^{iz}+e^{-iz}}=\dfrac{i}{2}\)

  2. Loser66
    • one year ago
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    hey, faking girl, help me out. hehehe

  3. dan815
    • one year ago
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    |dw:1444248216913:dw|

  4. Loser66
    • one year ago
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    \(e^{2iz}=1/3\)

  5. dan815
    • one year ago
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    |dw:1444248414375:dw|

  6. Loser66
    • one year ago
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    from above \(2(e^{iz} -e^{-iz}) = -(e^{iz}+e^{-iz})\) \(2e^{iz}-2e^{-iz}+e^{iz}+e^{-iz}=0\) \(3e^{iz}-e^{-iz}= 0\)

  7. Loser66
    • one year ago
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    multiple both sides by \(e^{iz}\), we get \(3e^{2iz} -1=0\\e^{2iz} = 1/3\)

  8. dan815
    • one year ago
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    |dw:1444248554037:dw|

  9. Loser66
    • one year ago
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    ok, you do your work, I do mine. then compare, hehehe. it's fun

  10. dan815
    • one year ago
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    |dw:1444248655592:dw|

  11. Loser66
    • one year ago
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    2iz = log (1/3) the RHS : \(log (1/3) = log|1/3| + i(arg (1/3) +2k\pi): k\in \mathbb Z\)

  12. dan815
    • one year ago
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    |dw:1444248766740:dw|

  13. Loser66
    • one year ago
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    now combine LHS \(z =\dfrac{log|1/3| + i arg(1/3+2k\pi)}{2i}=\dfrac{arg(1/3 + 2k\pi}{2} -\dfrac{ilog|1/3|}{2} \)

  14. Loser66
    • one year ago
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    You missed the real part!!!

  15. Loser66
    • one year ago
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    As what you draw, z has real part, right? is it not just imaginary.

  16. Loser66
    • one year ago
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    Yeah, you solve for y only, while z = x + iy.

  17. dan815
    • one year ago
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    =]

  18. dan815
    • one year ago
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    z is real in e^iz

  19. dan815
    • one year ago
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    e^itheta = cos theta + i sin theta there there is real

  20. dan815
    • one year ago
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    where theta* is real

  21. anonymous
    • one year ago
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    wow i must be dumb i dont get this at alll xD

  22. anonymous
    • one year ago
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    im 14 and in 11th grade but i dont get this

  23. anonymous
    • one year ago
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    You must teach me so i can master it

  24. anonymous
    • one year ago
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    Teach me

  25. anonymous
    • one year ago
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  26. Loser66
    • one year ago
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    hahaha.... I am not a teacher, how can I teach you, ask "Honor professor dan815" please

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