Complex no. Sum

[man111]

If [tex]z = e^{i\frac{8\pi}{11}}[/tex], Then Find [tex]Re(z+z^2+z^3+z^4+z^5)[/tex]

Where [tex]Re =[/tex] Real Part.

[martin123456]

I don't know if u know this: [tex]e^{ix}=\cos{x}+i\sin{x}[/tex]

[nikko]

Firstly for every [tex]x=e^{i\varphi}-1=-1+\cos\varphi+i\sin\varphi[/tex] it holds [tex]|x|=\sqrt{(-1{+}\cos\varphi)^2{+}\sin^2\varphi}=\sqrt{1+\cos^2\varphi-2\cos\varphi+\sin^2\varphi}=\sqrt{2-2\cos\varphi}=2\sin\frac{\varphi}{2}.[/tex]
So
[tex]x=2\sin\frac{\varphi}{2}\left(\frac{{-}1{+}\cos\varphi}{2\sin\frac{\varphi}{2}}+i\cos\frac{\varphi}{2}\right)=2\sin\frac{\varphi}{2}\left(-\sin\frac{\varphi}{2}+i\cos\frac{\varphi}{2}\right)=2\sin\frac{\varphi}{2}\left(\cos\left(\frac{\pi+\varphi}{2}\right)+i\sin\left(\frac{\pi+\varphi}{2}\right)\right)[/tex]

Now [tex]z=e^{\frac{8\pi}{11}}=\cos\frac{8\pi}{11}+i\sin\frac{8\pi}{11}[/tex], so
[tex]z+z^2+\dots+z^5=\frac{z^6-1}{z-1}-1=\frac{\cos\frac{48\pi}{11}+i\sin\frac{48\pi}{11}-1}{\cos\frac{8\pi}{11}+i\sin\frac{8\pi}{11}-1}-1=\frac{\cos\frac{4\pi}{11}+i\sin\frac{4\pi}{11}-1}{\cos\frac{8\pi}{11}+i\sin\frac{8\pi}{11}-1}-1=[/tex]
[tex]=\frac{2\sin\frac{2\pi}{11}\left(\cos\left(\frac{\pi+\frac{4\pi}{11}}{2}\right)+i\sin\left(\frac{\pi+\frac{4\pi}{11}}{2}\right)\right)}{2\sin\frac{4\pi}{11}\left(\cos\left(\frac{\pi+\frac{8\pi}{11}}{2}\right)+i\sin\left(\frac{\pi+\frac{8\pi}{11}}{2}\right)\right)}-1=\frac{\sin\frac{2\pi}{11}}{\sin\frac{4\pi}{11}}EXP(\frac{\pi+\frac{4\pi}{11}}{2}-\frac{\pi+\frac{8\pi}{11}}{2})-1=
\frac{\sin\frac{2\pi}{11}}{\sin\frac{4\pi}{11}}e^{\frac{2\pi}{11}}-1=[/tex]
[tex]=\frac{\sin\frac{2\pi}{11}}{\sin\frac{4\pi}{11}}(\cos\frac{2\pi}{11}+i\sin\frac{2\pi}{11})-1[/tex] so the real part is [tex]\text{Re}(z+\dots+z^6)=\frac{\sin\frac{2\pi}{11}\cos\frac{2\pi}{11}}{\sin\frac{4\pi}{11}}-1=\frac{\frac{1}{2}\sin\frac{4\pi}{11}}{\sin\frac{4\pi}{11}}-1=-\frac{1}{2}[/tex]