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