anonymous
  • anonymous
Find all complex numbers z satisfying...
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[z^3 = -27i\] Answers are: \[z = 3i\] \[z = \frac{ -3\sqrt{3} }{ 2 } - \frac{ 3i }{ 2 }\] \[z = \frac{ 3\sqrt{3} }{ 2 } - \frac{ 3i }{ 2 }\]
mayankdevnani
  • mayankdevnani
http://au.answers.yahoo.com/question/index?qid=20100318052317AAjz850
mayankdevnani
  • mayankdevnani

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anonymous
  • anonymous
|dw:1353827474036:dw|
anonymous
  • anonymous
how did you get pi/6?
anonymous
  • anonymous
i cant read the stuff at the top...
sirm3d
  • sirm3d
\[\huge -i = e^{(3\pi/2 + 2n\pi)}\]\[\huge z^3 = 3^3e^{(3\pi/2 + 2n\pi)}\]
sirm3d
  • sirm3d
\[\huge z=\sqrt[3]{3^3}e^{(\large \frac{ 3\pi }{ 2 }+2n \pi)/3},\text{ } n=0,1,2\]
anonymous
  • anonymous
how did you set up your first equation?
anonymous
  • anonymous
the 3pi/2
sirm3d
  • sirm3d
\[\huge z=3e^{\frac{ 3 \pi }{ 6 }+\frac{ 2 \pi }{ 3 }n},\text { }n=0,1,2\]
sirm3d
  • sirm3d
|dw:1353828234574:dw|
anonymous
  • anonymous
can you please explain your steps
anonymous
  • anonymous
for example, the 3pi/2 in the brackets
sirm3d
  • sirm3d
it's in the unit circle drawn above
anonymous
  • anonymous
lemme try it
anonymous
  • anonymous
do you use eulers equation? z = r(cos(angle) + i sin(angle))
sirm3d
  • sirm3d
yes, just slightly different notation \[re^{x i}=r(\cos x + i \sin x)\] |dw:1353828628571:dw|
anonymous
  • anonymous
thank you, it works!
anonymous
  • anonymous
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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