Question

If ${(2+z)}^6+{(2-z)}^6=0$ and $\omega =\frac{2+z}{2-z}$

• A. $\omega =e^i\frac{(2p+1)\pi }{6},p=0,1,2,3,4,5$
• B. $z=\frac{2(\omega -1)}{\omega +1}$
• C. $\omega =(-1{)}^{(}\frac{1}{6})$
• D. All of these

Answer 1

The correct option is D All of these

${(2+z)}^6{+(2-z)}^6=0\&w=\frac{2+z}{2-z}$
$\Rightarrow {\left(\frac{2+z}{2-z}\right)}^6=-1\Rightarrow w^6=-1$
${\therefore w=(-1)}^{\frac{1}{6}}$
$\because \frac{2+z}{2-z}=w\Rightarrow 2(w-1)=z(w+1)$
$\therefore z=\frac{2(w-1)}{w+1}$
${\because w=(-1)}^{\frac{1}{6}}$
$w={(\cos \pi +i\sin \pi )}^{\frac{1}{6}}=\cos \left(\frac{2p\pi +\pi }{6}\right)+i\sin \left(\frac{2p\pi +\pi }{6}\right)$ ..{De Moivre's Theorem}

Where$p=0,1,2,3,4,5.$
$\Rightarrow w=e^{i\frac{(2p+1)\pi }{6}}$
Hence, option 'D' is correct.