Question

Proove that the roots of the equation $(z+1{)}^6+(z-1{)}^6=0$ are given by $\pm cot\frac{\pi }{12},\pm cot\frac{3\pi }{12},\pm cot\frac{5\pi }{12}$

Answer 1

We have, $\frac{z+1}{z-1}=(-1{)}^{1/6}=(cos\pi \pm isin\pi {)}^{1/6}$
Or $\frac{z+1}{z-1}=(cos\frac{(2n\pi +\pi )}{6}\pm isin\frac{(2n\pi +\pi )}{6})$
where n = 0, 1, 2
Or $\frac{z+1}{z-1}=\frac{cos\theta \pm isin\theta }{i}where\theta =(2n+1)\frac{\pi }{6}$
Apply componendo and dividendo,
$\frac{2z}{2}=\frac{(1+cos\theta )\pm isin\theta }{(cos\theta -1)\pm isin\theta }$
$z=\frac{2co{s}^2\frac{\theta }{2}\pm i2sin\frac{\theta }{2}cos\frac{\theta }{2}}{-2si{n}^2\frac{\theta }{2}\pm i2sin\frac{\theta }{2}cos\frac{\theta }{2}}$
$z=\frac{2cos\frac{\theta }{2}(cos\frac{\theta }{2}\pm i2sin\frac{\theta }{2})}{\pm 2sin\frac{\theta }{2}(cos\frac{\theta }{2}\pm i2sin\frac{\theta }{2})}\because -1=i^2$
Or z = $\pm icot\frac{\theta }{2}=\pm cot(2n+1)\frac{\pi }{12}$
where n = 0, 1, 2
Or z = $\pm icot\frac{\pi }{12},\pm cot\frac{3\pi }{12},\pm cot\frac{5\pi }{12}$