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iop360

  • 2 years ago

Find all complex numbers z satisfying...

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  1. iop360
    • 2 years ago
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    \[z^3 = -27i\] Answers are: \[z = 3i\] \[z = \frac{ -3\sqrt{3} }{ 2 } - \frac{ 3i }{ 2 }\] \[z = \frac{ 3\sqrt{3} }{ 2 } - \frac{ 3i }{ 2 }\]

  2. mayankdevnani
    • 2 years ago
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    http://au.answers.yahoo.com/question/index?qid=20100318052317AAjz850

  3. mayankdevnani
    • 2 years ago
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    ok @iop360

  4. mahmit2012
    • 2 years ago
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    |dw:1353827474036:dw|

  5. iop360
    • 2 years ago
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    how did you get pi/6?

  6. iop360
    • 2 years ago
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    i cant read the stuff at the top...

  7. sirm3d
    • 2 years ago
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    \[\huge -i = e^{(3\pi/2 + 2n\pi)}\]\[\huge z^3 = 3^3e^{(3\pi/2 + 2n\pi)}\]

  8. sirm3d
    • 2 years ago
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    \[\huge z=\sqrt[3]{3^3}e^{(\large \frac{ 3\pi }{ 2 }+2n \pi)/3},\text{ } n=0,1,2\]

  9. iop360
    • 2 years ago
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    how did you set up your first equation?

  10. iop360
    • 2 years ago
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    the 3pi/2

  11. sirm3d
    • 2 years ago
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    \[\huge z=3e^{\frac{ 3 \pi }{ 6 }+\frac{ 2 \pi }{ 3 }n},\text { }n=0,1,2\]

  12. sirm3d
    • 2 years ago
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    |dw:1353828234574:dw|

  13. iop360
    • 2 years ago
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    can you please explain your steps

  14. iop360
    • 2 years ago
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    for example, the 3pi/2 in the brackets

  15. sirm3d
    • 2 years ago
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    it's in the unit circle drawn above

  16. iop360
    • 2 years ago
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    lemme try it

  17. iop360
    • 2 years ago
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    do you use eulers equation? z = r(cos(angle) + i sin(angle))

  18. sirm3d
    • 2 years ago
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    yes, just slightly different notation \[re^{x i}=r(\cos x + i \sin x)\] |dw:1353828628571:dw|

  19. iop360
    • 2 years ago
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    thank you, it works!

  20. iop360
    • 2 years ago
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    only 1 question though, for your first equation -i = ... if you have a negative i, does that always mean 3pi/2

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