tożsamość z cosinusem

[mol_ksiazkowy]

Udowodnić tę tożsamość \(\displaystyle{ z^{2n+1}+1 =(z+1) \prod_{k=1}^{n} (z^2+2z \cos\frac{2k\pi}{2n+1} +1)}\)

[kerajs]

\(\displaystyle{ z^{2n+1}+1 = (z-z _{0} )(z-z _{1} )(z-z _{2} ) \cdot .... \cdot (z-z _{2n} )=.....}\)

Pierwiastki z postaci iloczynowej to rozwiązania równania:
\(\displaystyle{ z^{2n+1}=-1 \\ z^{2n+1}=1\left( \cos ( \pi +k2 \pi )+i \sin ( \pi +k2 \pi ) \right)}\)
\(\displaystyle{ z= \sqrt[2n+1]{\cos ( \pi +k2 \pi )+i \sin ( \pi +k2 \pi )}= \cos \frac{ \pi +k2 \pi }{2n+1}+i \sin \frac{ \pi +k2 \pi }{2n+1}}\)

\(\displaystyle{ z_1= \cos \frac{ \pi +0 \cdot 2 \pi }{2n+1}+i \sin \frac{ \pi +0 \cdot2 \pi }{2n+1}=}\)
\(\displaystyle{ z_2= \cos \frac{ \pi +1 \cdot2 \pi }{2n+1}+i \sin \frac{ \pi +1 \cdot2 \pi }{2n+1}}\)
....
\(\displaystyle{ z_j= \cos \frac{ \pi +j \cdot2 \pi }{2n+1}+i \sin \frac{ \pi +j \cdot2 \pi }{2n+1}}\)
....
\(\displaystyle{ z_{n}= \cos \frac{ \pi +(n-1) \cdot2 \pi }{2n+1}+i \sin \frac{ \pi +(n-1)\cdot2 \pi }{2n+1}}\)

\(\displaystyle{ z_0= \cos \frac{ \pi +n \cdot2 \pi }{2n+1}+i \sin \frac{ \pi +n\cdot2 \pi }{2n+1}=-1}\)

\(\displaystyle{ z_{n+1}= \cos \frac{ \pi +(n+1)\cdot2 \pi }{2n+1}+i \sin \frac{ \pi +(n+1)\cdot2 \pi }{2n+1}}\)
...
\(\displaystyle{ z_{2n}= \cos \frac{ \pi +2n\cdot2 \pi }{2n+1}+i \sin \frac{ \pi +2n\cdot2 \pi }{2n+1}}\)


Można też zauważyć że
\(\displaystyle{ z _{j}=\overline{z_{2n+1-j}}}\)

\(\displaystyle{ ....= (z+1 )(z-z _{1} )(z-z _{2} ) \cdot .... \cdot (z-z _{n} )(z-\overline{z_1} )(z-\overline{z_2} ) \cdot .... \cdot (z-\overline{z_n} )=\\=(z+1) \prod_{k=1}^{n}(z-z _{k} )(z-\overline{z_k} )=(z+1) \prod_{k=1}^{n}(z^2-z(z _{k} +\overline{z_k} )+z _{k} \overline{z_k} )=\\=(z+1) \prod_{k=1}^{n}(z^2-z 2 \Re (z _{k})+\left| z _{k}\right|^2 ) =(z+1) \prod_{k=1}^{n} (z^2 -2z \cos\frac{ \pi +2k\pi}{2n+1} +1) \neq \\ \neq
(z+1) \prod_{k=1}^{n} (z^2 +2z \cos\frac{2k\pi}{2n+1} +1)}\)

[a4karo]

\(\displaystyle{ ....= (z+1 )(z-z _{1} )(z-z _{2} ) \cdot .... \cdot (z-z _{n} )(z-\overline{z_1} )(z-\overline{z_2} ) \cdot .... \cdot (z-\overline{z_n} )=\\=(z+1) \prod_{k=1}^{n}(z-z _{k} )(z-\overline{z_k} )=(z+1) \prod_{k=1}^{n}(z^2-z(z _{k} +\overline{z_k} )+z _{k} \overline{z_k} )=\\=(z+1) \prod_{k=1}^{n}(z^2-z 2 \Re (z _{k})+\left| z _{k}\right|^2 ) =(z+1) \prod_{k=1}^{n} (z^2 -2z \cos\frac{ \pi +2k\pi}{2n+1} +1) \neq \\ \neq
(z+1) \prod_{k=1}^{n} (z^2 +2z \cos\frac{2k\pi}{2n+1} +1)}\)

Ale \(\displaystyle{ \cos\frac{ \pi +2(n-k)\pi}{2n+1}=-\cos\frac{ 2k\pi}{2n+1}}\), więc pewnie jednak wzór jest prawdziwy