rvc
  • rvc
complex numbers
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rvc
  • rvc
show that the roots of \(\huge (x-1)^5=32(x+1)^5 are~given~by~\\\huge x=\huge \frac{-3+4isin \huge\frac{2n\pi }{5}}{5-4cos\huge \frac{2n\pi}{5}}\)
rvc
  • rvc
@ganeshie8 @iambatman @triciaal @Callisto @Loser66 @Astrophysics @Directrix @bibby
Maddy1251
  • Maddy1251
@Owlcoffee

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More answers

Owlcoffee
  • Owlcoffee
Okay, I think that we should first calculate the roots for the function, they will for sure be complex.
rvc
  • rvc
yep
Owlcoffee
  • Owlcoffee
Then, you should be able to express those roots in their polar form.
rvc
  • rvc
hmm.. idk
lochana
  • lochana
can you provide where you got this problem from?
rvc
  • rvc
from my book
fouzberzerk
  • fouzberzerk
Hey Admir, I could still use a little help on a subject.
Michele_Laino
  • Michele_Laino
hint: we can rewrite your equation as follows: \[\huge {\left( {\frac{{z - 1}}{{z + 1}} \cdot \frac{1}{2}} \right)^5} = 1\] so we get: \[\Large \begin{gathered} \frac{{z - 1}}{{z + 1}} \cdot \frac{1}{2} = \sqrt[5]{1} = \cos \left( {\frac{{2\pi n}}{5}} \right) + i\sin \left( {\frac{{2\pi n}}{5}} \right),\quad \hfill \\ \hfill \\ n = 0,1,2,3,4 \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
please solve for \(\Large z\)
rvc
  • rvc
i have done till z after that im not getting
Michele_Laino
  • Michele_Laino
please wait, I'm computing...
Michele_Laino
  • Michele_Laino
I got this: \[\huge z = \frac{{1 + 2{e^{\frac{{2\pi ni}}{5}}}}}{{1 - 2{e^{\frac{{2\pi ni}}{5}}}}}\]
Michele_Laino
  • Michele_Laino
I think that we are in the right way, since I got your result
Michele_Laino
  • Michele_Laino
here are the next steps: \[\Large \begin{gathered} z = \frac{{1 + 2\cos \theta + 2i\sin \theta }}{{1 - 2\cos \theta - 2i\sin \theta }} = \hfill \\ \hfill \\ = \frac{{\left( {1 + 2\cos \theta } \right) + 2i\sin \theta }}{{\left( {1 - 2\cos \theta } \right) - 2i\sin \theta }} \cdot \frac{{\left( {1 - 2\cos \theta } \right) + 2i\sin \theta }}{{\left( {1 - 2\cos \theta } \right) + 2i\sin \theta }} = ...? \hfill \\ \end{gathered} \] where: \[\Large \theta = \frac{{2\pi n}}{5}\]
rvc
  • rvc
hmmmmmmmm
jango_IN_DTOWN
  • jango_IN_DTOWN
@rvc
rvc
  • rvc
thanks @Michele_Laino :)

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