Question

If $n >1$, then the roots of $z^{n}=(z+1)^{n}$ lie on a

• A. circle
• B. straight line
• C. parabola
• D. None of these

Answer 1

Answer: (B)

We have, $z^{n}=(z+1)^{n}$
$\Rightarrow \left(\frac{z+1}{z}\right)^{n}=1=\cos 0+i \sin 0 $
$\Rightarrow \frac{z+1}{z}=(\cos 2 \pi r+i \sin 2 \pi r)^{1 / n} $
$=\cos \frac{2 \pi r}{n}+i \sin \frac{2 \pi r}{n}$
where, $r=0,1,2, \ldots, n-1$.
$\Rightarrow 1+\frac{1}{z}=\cos \frac{2 \pi r}{n}+i \sin \frac{2 \pi r}{n}$
$\Rightarrow 1+\frac{1}{z}=1-2 \sin ^{2} \frac{r \pi}{n}+i \times 2 \sin \frac{\pi r}{n} \times \cos \frac{\pi r}{n}$
$\Rightarrow \frac{1}{z}=-2 \sin ^{2} \frac{r \pi}{n}+2 i \times \sin \frac{r \pi}{n} \cos \frac{r \pi}{n}$
$\Rightarrow z=\frac{1}{i\left(2 \sin \frac{r \pi}{n}\right)\left[\cos \frac{r \pi}{n}+i \sin \frac{r \pi}{n}\right]}$
$=\frac{1}{i\left(2 \sin \frac{r \pi}{n}\right)}\left[\cos \frac{r \pi}{n}-i \sin \frac{r \pi}{n}\right]$
$\Rightarrow x+i y=-\frac{i}{2} \cot \frac{r \pi}{n}-\frac{1}{2} $
$\Rightarrow x=-\frac{1}{2}$
Hence, all the points lie on the line $x=-\frac{1}{2}$