Question

complex numbers

Answer 1

show that the roots of \(\huge (x-1)^5=32(x+1)^5 are~given~by~\\\huge x=\huge \frac{-3+4isin \huge\frac{2n\pi }{5}}{5-4cos\huge \frac{2n\pi}{5}}\)

Answer 2

@ganeshie8 @iambatman @triciaal @Callisto @Loser66 @Astrophysics @Directrix @bibby

Answer 3

@Owlcoffee

Answer 4

Okay, I think that we should first calculate the roots for the function, they will for sure be complex.

Answer 5

yep

Answer 6

Then, you should be able to express those roots in their polar form.

Answer 7

hmm..
idk

Answer 8

can you provide where you got this problem from?

Answer 9

from my book

Answer 10

Hey Admir, I could still use a little help on a subject.

Answer 11

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:
\[\Large \begin{gathered}
\frac{{z - 1}}{{z + 1}} \cdot \frac{1}{2} = \sqrt[5]{1} = \cos \left( {\frac{{2\pi n}}{5}} \right) + i\sin \left( {\frac{{2\pi n}}{5}} \right),\quad \hfill \\
\hfill \\
n = 0,1,2,3,4 \hfill \\
\end{gathered} \]

Answer 12

please solve for \(\Large z\)

Answer 13

i have done till z
after that im not getting

Answer 14

please wait, I'm computing...

Answer 15

I got this:
\[\huge z = \frac{{1 + 2{e^{\frac{{2\pi ni}}{5}}}}}{{1 - 2{e^{\frac{{2\pi ni}}{5}}}}}\]

Answer 16

I think that we are in the right way, since I got your result

Answer 17

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:
\[\Large \theta = \frac{{2\pi n}}{5}\]

Answer 18

hmmmmmmmm

Answer 19

@rvc

Answer 20

thanks @Michele_Laino :)