Question

If $\alpha =\cos \left(\frac{8\pi }{11}\right)+i\sin \left(\frac{8\pi }{11}\right)$ then $Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

$-\frac{1}{2}$

Given

$\alpha =\cos \frac{8\pi }{11}+i\sin \frac{8\pi }{11}=e^{\frac{i8\pi }{11}}{\alpha }^2=e^{\frac{i16\pi }{11}}=\cos \frac{16\pi }{11}+i\sin \frac{16\pi }{11}{\alpha }^3=e^{\frac{i24\pi }{11}}=\cos \frac{24\pi }{11}+i\sin \frac{24\pi }{11}{\alpha }^4=e^{\frac{i32\pi }{11}}=\cos \frac{32\pi }{11}+i\sin \frac{32\pi }{11}{\alpha }^5=e^{\frac{i40\pi }{11}}=\cos \frac{40\pi }{11}+i\sin \frac{40\pi }{11}$

Now,

$Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)=\cos \frac{8\pi }{11}+\cos \frac{16\pi }{11}+\cos \frac{24\pi }{11}+\cos \frac{32\pi }{11}+\cos \frac{40\pi }{11}$

Say $x=\frac{8\pi }{11}$,then

$Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)=\cos x+\cos 2x+\cos 3x+\cos 4x+\cos 5x$

Multiply and divide by $2\sin \frac{x}{2}$

$Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)=\frac{1}{2\sin \frac{x}{2}}(2\sin \frac{x}{2}\cos x+2\sin \frac{x}{2}.\cos 2x+2\sin \frac{x}{2}.\cos 3x+2\sin \frac{x}{2}.\cos 4x+2\sin \frac{x}{2}.\cos 5x)$

$=\frac{1}{2\sin \frac{x}{2}}[\sin \frac{3x}{2}-\sin \frac{x}{2}+\sin \frac{5x}{2}-\sin \frac{3x}{2}+\sin \frac{7x}{2}-\sin \frac{5x}{2}+\sin \frac{9x}{2}-\sin \frac{7x}{2}+\sin \frac{11x}{2}-\sin \frac{9x}{2}]=\frac{1}{2\sin \frac{x}{2}}[\sin \frac{11x}{2}-\sin \frac{x}{2}]$

Putting $x=\frac{8\pi }{11}$

$Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)=\frac{1}{2\sin \frac{4\pi }{11}}[\sin 4\pi -\sin \frac{4\pi }{11}]=\frac{-1}{2}$