Question

If $z=\cos \frac{8\pi }{11}+i\sin \frac{8\pi }{11},$ then Real $(z+z^2+z^3+z^4+z^5)$ is

• A. $-\frac{1}{2}$
• B. 0
• C. $\frac{1}{2}$
• D. none

Answer 1

The correct option is A $-\frac{1}{2}$

Real (z) $=\frac{z+\overline{z}}{2}=\frac{1}{2}(z+\frac{1}{z})$

$\because z\overline{z}={\cos }^2\frac{8\pi }{11}+{\sin }^2\frac{8\pi }{11}=1\therefore \overline{z}=\frac{1}{z}$

$\because$ E=Real part of $(z+z^2+z^3+z^4+z^5+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+\frac{1}{z^5})$

Now $z^{11}=\cos 8\pi +i\sin 8\pi =1$

$\therefore \frac{1}{z^4}=\frac{z^7}{z^{11}}=z^7$ etc.
$\therefore E=\frac{1}{2}[z+z^2+z^3+z^4+z^5+z^{10}+z^9+z^8+z^7+z^6]$

Add and subtract $z^{11}.$
$\therefore E=\frac{1}{2}$ [sum of G.P. of 11 terms-$z^{11}$]

$=\frac{1}{2}[\frac{z(1-z^{11})}{1-z}-z^{11}]=\frac{1}{2}(0-1)=-\frac{1}{2}$ by (I)