# complex numbers problem

• Apr 22nd 2013, 03:50 AM
mathfag188
complex numbers problem
the question and solution are attached but i cant understand, can somebody please guide
• Apr 22nd 2013, 04:22 AM
chiro
Re: complex numbers problem
Hey mathfag188.

Are you aware of De-Moivre's theorem and how to use that to solve roots for z^n = 1?
• Apr 22nd 2013, 04:35 AM
Prove It
Re: complex numbers problem
\displaystyle \displaystyle \begin{align*} \left( \frac{z +1 }{z} \right) ^5 &= 1 \\ \left( \frac{z + 1}{z} \right) ^5 &= e^{2\pi \, i \, k} \textrm{ where } k \in \mathbf{Z} \\ \frac{z + 1}{z} &= e^{\frac{2\pi \, i \, k}{5}} \\ z + 1 &= e^{\frac{2\pi\,i\,k}{5}}\,z \\ 1 &= \left( e^{\frac{2\pi \, i \, k}{5}} - 1 \right) z \\ \frac{1}{ e^{\frac{2\pi \, i \, k}{5}} - 1} &= z \\ \frac{1}{\cos{\left( \frac{2\pi \, k}{5} \right) } + i\sin{\left( \frac{2\pi\,k}{5} \right) } -1 } &= z \\ \frac{\cos{\left( \frac{2\pi\,k}{5} \right) } - 1 - i\sin{\left( \frac{2\pi\,k}{5} \right) } }{\left[ \cos{ \left( \frac{2\pi\,k}{5} \right) } - 1 + i\sin{ \left( \frac{2\pi\,k}{5} \right) } \right] \left[ \cos{ \left( \frac{2\pi\,k}{5} \right) } - 1 - i\sin{ \left( \frac{2\pi\,k}{5} \right) } \right] } &= z \end{align*}

\displaystyle \displaystyle \begin{align*} \frac{\cos{ \left( \frac{2\pi\,k}{5} \right) } - 1 - i\sin{ \left( \frac{2\pi\,k}{5} \right) }}{ \left[ \cos{ \left( \frac{2\pi\,k}{5} \right) } - 1 \right] ^2 + \sin^2{ \left( \frac{2\pi\,k}{5} \right) } } &= z \\ \frac{ \cos{ \left( \frac{2\pi\,k}{5} \right) } - 1 - i\sin{ \left( \frac{2\pi\,k}{5} \right) } }{ \cos^2{ \left( \frac{2\pi\,k}{5} \right) } - 2\cos{ \left( \frac{2\pi\,k}{5} \right) } + 1 + \sin^2{ \left( \frac{2\pi\,k}{5} \right) } } &= z \\ \frac{ \cos{ \left( \frac{2\pi\,k}{5} \right) } - 1 - i\sin{ \left( \frac{2\pi\,k}{5} \right) } }{ 2 \left[ 1 - \cos{\left( \frac{2\pi\,k}{5} \right) } \right] } &= z \\ \frac{1}{2} \left\{ -1 + \left[ \frac{\sin{\left( \frac{2\pi\,k}{5} \right)}}{1 - \cos{\left( \frac{2\pi\,k}{5} \right)}} \right] i \right \} &= z \end{align*}

\displaystyle \displaystyle \begin{align*} \frac{1}{2} \left( -1 + \left\{ \frac{\sin{ \left( \frac{2\pi\,k}{5} \right) }\left[ 1 + \cos{ \left( \frac{2\pi\,k}{5} \right) } \right]}{\left[ 1 - \cos{ \left( \frac{2\pi\,k}{5} \right) } \right] \left[ 1 + \cos{ \left( \frac{2\pi\,k}{5} \right) } \right] } \right\} i \right) &= z \\ \frac{1}{2} \left( -1 + \left\{ \frac{\sin{ \left( \frac{2\pi\,k}{5} \right) }\left[ 1 + \cos{ \left( \frac{2\pi\,k}{5} \right) } \right]}{ 1 - \cos^2{ \left( \frac{2\pi\,k}{5} \right) } } \right\} i \right) &= z\\ \frac{1}{2} \left( -1 + \left\{ \frac{\sin{ \left( \frac{2\pi\,k}{5} \right) }\left[ 1 + \cos{ \left( \frac{2\pi\,k}{5} \right) } \right]}{ \sin^2{ \left( \frac{2\pi\,k}{5} \right) } } \right\} i \right) &= z \\ \frac{1}{2} \left\{ -1 + \left[ \frac{ 1 + \cos{ \left( \frac{2\pi\,k}{5} \right) } }{ \sin{ \left( \frac{2\pi\,k}{5} \right) } } \right] i \right\} &= z \end{align*}
• Apr 22nd 2013, 05:48 AM
mathfag188
Re: complex numbers problem
I dont understand what yuo did in 8th step, how you expanded it?
other thing, did you read the solution i posted? there its done in almost 5 steps :P

actually what i wanted to ask and i was stuck on this step
how does the following happen?
in solution it says "cis" where i have put cos, as i was thinking it was a typo in solution but if there is something as cis do let me know(Worried)

$\displaystyle z=\frac{-1}{1-cos(\frac{2k\pi}{5})} = \frac{-(cos(-\frac{-k\pi}{5}))}{cos(-\frac{k\pi}{5})-cos(\frac{k\pi}{5})}$
• Apr 22nd 2013, 06:07 AM
Prove It
Re: complex numbers problem
Surely you know how to expand two binomials...

You're welcome btw ><
• Apr 22nd 2013, 06:26 AM
mathfag188
Re: complex numbers problem
but my problem isnt solved yet