iop360
Find all complex numbers z satisfying...



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iop360
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\[z^3 = 27i\]
Answers are:
\[z = 3i\]
\[z = \frac{ 3\sqrt{3} }{ 2 }  \frac{ 3i }{ 2 }\]
\[z = \frac{ 3\sqrt{3} }{ 2 }  \frac{ 3i }{ 2 }\]


mayankdevnani
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ok @iop360

mahmit2012
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dw:1353827474036:dw

iop360
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how did you get pi/6?

iop360
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i cant read the stuff at the top...

sirm3d
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\[\huge i = e^{(3\pi/2 + 2n\pi)}\]\[\huge z^3 = 3^3e^{(3\pi/2 + 2n\pi)}\]

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

iop360
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how did you set up your first equation?

iop360
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the 3pi/2

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

sirm3d
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dw:1353828234574:dw

iop360
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can you please explain your steps

iop360
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for example, the 3pi/2 in the brackets

sirm3d
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it's in the unit circle drawn above

iop360
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lemme try it

iop360
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do you use eulers equation?
z = r(cos(angle) + i sin(angle))

sirm3d
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yes, just slightly different notation
\[re^{x i}=r(\cos x + i \sin x)\]
dw:1353828628571:dw

iop360
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thank you, it works!

iop360
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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