Question

Solve the equation $(1+z{)}^8+z^8=0$ and shown that real part of each root of the equation is $-\frac{1}{2}$.

Answer 1

Let $(1+z{)}^8+z^8=0$

or,$\frac{1+z}{z}=(-1{)}^{1/8}=(\cos \pi +i\sin \pi {)}^{1/8}$
or, $\frac{1}{z}+1=\cos \frac{(2n\pi +\pi )}{8}+i\sin \frac{(2n\pi +\pi )}{8}$
or, $\frac{1}{z}=-1+\cos \theta +i\sin \theta$
where $\theta$ = (2n + 1) $\frac{\pi }{8}$
$\frac{1}{z}=-2{\sin }^2\frac{\theta }{2}+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2},put-1=i^2$
$=+i2\sin \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})$
$\therefore z=\frac{1}{+2i\sin \left(\theta /2\right)}(\cos \frac{\theta }{2}+-i\sin \frac{\theta }{2})$
$=\frac{1}{2}(-1-i\cot \frac{\theta }{2})$
$\because \frac{1}{i}=-i$
where $\frac{\theta }{2}=(2n+1)\frac{\pi }{16}$ and $n$ varies from $0$ to $7$.
Clearly real part of every roots is $-\frac{1}{2}$