21is the answr -0.5
Let alpha = e ^ i8pi / 11 , then find Re ( alpha + alpha ^2 + alpha ^3 + alpha ^4 + alpha ^5).
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21actually its easy
(cos∂+ i sin∂) + (cos∂+ i sin∂)2 + (cos∂+ i sin∂)n=
{1 - cos(n+1)∂- i sin(n+1)∂}/{1-cos∂ - i sin∂) - 1 .........1st
(GP summation and De Moivres theorem)
putting ∂= 8∩/11 and n=5 we get
Re(a + a2 + ....+ a5) = real part of RHS in 1st equation
which comes to be {sin(88∩/22) - sin 4∩/11}/2 sin 4∩/11
so the answer is -0.5
1708$Here $\alpha = e^{i\frac{8\pi}{11}}=cos(\frac{8\pi}{11})+isin(\frac{8\pi}{11})$\\\\ Similarly $\alpha^2=e^{i\frac{16\pi}{11}}=cos(\frac{16\pi}{11})+isin(\frac{16\pi}{11})$\\\\ In a Similar manner $\alpha^3=cos(\frac{24\pi}{11})+isin(\frac{24\pi}{11})$\\\\ $\alpha^4=cos(\frac{32\pi}{11})+isin(\frac{32\pi}{11})$ and $\alpha^5=cos(\frac{40\pi}{11})+isin(\frac{40\pi}{11})$\\\\ So $Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)$\\\\ $=cos(\frac{8\pi}{11})+cos(\frac{16\pi}{11})+cos(\frac{24\pi}{11})+cos(\frac{32\pi}{11})+cos(\frac{40\pi}{11})$\\\\ Now Let $x=\frac{8\pi}{11}$, and $S=cosx+cos2x+cos3x+cos4x+cos5x$\\\\\ $S=\frac{1}{2sin\frac{x}{2}}(2sin\frac{x}{2}cosx+2sin\frac{x}{2}cos2x+2sin\frac{x}{2}cos3x+2sin\frac{x}{2}cos4x+2sin\frac{x}{2}cos5x)$\\\\ After Simplifying We get\\\\ $S=\frac{1}{2sin\frac{x}{2}}(sin\frac{11x}{2}-sin\frac{x}{2})=-\frac{1}{2}$\\\\\\\\ bcz $sin(\frac{11x}{2})=sin(\frac{11\times 8\pi}{2\times 11})=0$\\\\
$So $\boxed{\boxed{Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)=-\frac{1}{2}}}$
341Remember that if \alpha = e^{\frac{i2 \pi}{n}}, and \beta = e^{\frac{i2k \pi}{n}} where gcd (k,n)=1, then
\beta^k ; 1 \le k \le 10 generates all the 10 non-real roots of z^{11}=1 i.e. all
10 roots of z^{10}+z^9+...+z^2+z+1=0
That means the roots are \beta, \overline{\beta}, \beta^2, \overline{\beta^2},.., \beta^5, \overline{\beta^5}
So, the sum of roots =
\sum_{i=1}^5 \left(\beta^i + \overline{\beta^i} \right)= 2\sum_{i=1}^5 \Re \left(\beta^i \right) = -1 \Rightarrow \sum_{i=1}^5 \Re \left(\beta^i \right)=-\frac{1}{2}
