Question

If $z=e^{\frac{i\pi }{13}}$ then $\frac{1}{1-z}$ is equal to $(i=\sqrt{-1})$

• A. $1-z^2+z^4-z^6+z^8-z^{10}+z^{12}$
• B. $1+z^2+z^4+z^6+z^8+z^{10}+z^{12}$
• C. $z-z^3+z^5-z^7+z^3-z^{11}$
• D. $z+z^3+z^5+z^7+z^5+z^{11}$

Answer 1

Answer: (B), (D)

$z+z^3+z^5+z^7+z^5+z^{11}$

$z=e^{i\frac{\pi }{13}}\Rightarrow z^{13}=-1$

$z^{13}+1=0\Rightarrow (z+1)(z^{12}-z^{11}+\dots \dots \dots +1)=0$

$\Rightarrow z^{12}-z^{11}+\dots \dots \dots z+1=0$ ........ (1)

$1-z^{13}=2$

$\Rightarrow \frac{1-z^{13}}{1-z}=\frac{2}{1-z}$

$\Rightarrow 1+z+z^2+\dots \dots \dots +z^{12}=\frac{2}{1-z}$ ....... (2)

From (1) and (2)

$\begin{array}{c}1+z^2+z^4+\dots \dots \dots +z^{12}=\frac{1}{1-z}\\ z+z^3+\dots \dots \dots +z^{11}=\frac{1}{1-z}\end{array}\}$