Didee
  • Didee
a solution for z^3 +2 + 2 sqrt3i = 0, by using the n^th root formula for complex numbers is?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
go to this link : http://en.wikipedia.org/wiki/Square_root
Didee
  • Didee
Thanks, but its not helping much
anonymous
  • anonymous
express the complex number in the form: \[e ^{i \theta}\]

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anonymous
  • anonymous
using eulers formula we have \[R e^{i\theta} =R( \cos\theta + isin\theta) \] hence \[(R e^{i\theta})^n = R^n e^{i n\theta} = R^n(cosn\theta + isinn\theta)\] in your case n = 3 so express z^3 as \[R^3(cos3\theta + isin3\theta) = -2 -2\sqrt{3}i \] now solve for R and theta
anonymous
  • anonymous
taking the modulus of both sides: \[R^3 = \sqrt{2^2 + (2\sqrt{3})^2}\]
anonymous
  • anonymous
and finding the argument: \[tan3\theta = \frac{-2\sqrt{3}}{-2} = \sqrt{3}\]
anonymous
  • anonymous
do you follow?
Didee
  • Didee
thanks, but I'm completely clueless... think I need some background, any suggestions? I have possible answers, see attached
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anonymous
  • anonymous
what background knowledge do you have on this already?
anonymous
  • anonymous
complex numbers i mean
Didee
  • Didee
not much hey, basically what a complex number is and some formulas to use: |a+bi| = \[\sqrt{a^2+b^2}\] a=r cos \[\theta\] b = r sin \[\theta\] a + bi =r\[\left( \cos \theta + i \sin \theta \right)\]
anonymous
  • anonymous
ok, have you seen the geometrical effect of complex multiplication? try this: let\[z_1 = 1+i \text{ , }z_2 = 2+ \sqrt{3}i\] find the argument and modulus of each z and then find the argument and modulus of \[z_1z_2 = (1+i)(2+\sqrt{3}i)\] see if you can work out the relationship
anonymous
  • anonymous
if you have come across compound angle formulae, try doing this in general with: \[z_1 = \cos\theta + i \sin\theta \text{ , } z_2 = \cos\phi + i \sin\phi\]
anonymous
  • anonymous
remember all the while what you are doing geometrically:|dw:1333620316032:dw|
Didee
  • Didee
ok, give me some time to work on this, will reply a bit later...
anonymous
  • anonymous
sure :) i love this area of maths, where complex and trig overlap
Didee
  • Didee
maybe I'll get there one day:-)
Didee
  • Didee
Hi, still couldn't figure this one out so will tackle it some time later again. Thanks anyway.

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