Question

If $z=x+iy(x,y\in R,x\ne -1/2),$ then the number of values of $z$ satisfying ${\left|z\right|}^n=z^2{\left|z\right|}^{n-2}+z{\left|z\right|}^{n-2}+1$ for all values of $n$ $(n\in N,n>1)$, is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

Let $z=r{e}^{i\alpha }$.
The given equation can be rewritten as -
$r^n=r^n\left(e^{2i\alpha }\right)+r^{n-1}\left(e^{i\alpha }\right)+1$
Equating the real and imaginary parts we get,
$r^n-1=r^n\left(cos2\alpha \right)+r^{n-1}\left(cos\alpha \right)$ -(i) and
$r^n\left(sin2\alpha \right)+r^{n-1}\left(sin\alpha \right)=0$ - (ii)
(ii) $\Rightarrow sin\alpha =0$ or $1+2rcos\alpha =0$
Let us consider the first case, $sin\alpha =0$.
Hence, $cos\alpha =1$ or $cos\alpha =-1$.
If $cos\alpha =1$,
(i) $\Rightarrow r^n+r^{n-1}=r^n-1$
$\Rightarrow r^{n-1}=-1$. This is not possible.
If $cos\alpha =-1$,
(i)$\Rightarrow r^n-r^{n-1}=r^n-1$
$\Rightarrow r^{n-1}=1$
$\Rightarrow r=1$
Hence, $z=-1$.
Let us consider the second case :- $1+2rcos\alpha =0$
$\Rightarrow cos\alpha =\frac{-1}{2r}$
(i) $\Rightarrow r^n(\frac{1}{2r^2}-1)+r^{n-1}\left(\frac{-1}{2r}\right)=r^n-1$
On simplifying we get,
$r^n=\frac{1}{2}$
However, this cannot be true for all values of $n$.
Hence, there is only one solution