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

The given equation is
$\begin{array}{l}|\mathrm{z}{|}^{\mathrm{n}}=\left({\mathrm{z}}^2+\mathrm{z}\right)|\mathrm{z}{|}^{\mathrm{n}-2}+1\\ \Rightarrow {\mathrm{z}}^2+\mathrm{z}\text{ is real }\\ \Rightarrow {\mathrm{z}}^2+\mathrm{z}={\overline{\mathrm{z}}}^2+\overline{\mathrm{z}}\\ \Rightarrow (\mathrm{z}-\overline{\mathrm{z}})(\mathrm{z}+\overline{\mathrm{z}}+1)=0\\ \Rightarrow \mathrm{z}=\overline{\mathrm{z}}=\mathrm{x}\text{ as }\mathrm{z}+\overline{\mathrm{z}}+1\ne 0(\mathrm{x}\ne -1/2)\end{array}$
Hence, the given equation reduces to
$\begin{array}{l}{\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{n}}+\mathrm{x}|\mathrm{x}{|}^{\mathrm{n}-2}+1\\ \Rightarrow \mathrm{x}|\mathrm{x}{|}^{\mathrm{n}-2}=-1\\ \Rightarrow \mathrm{x}=-1\end{array}$
So, the number of solutions is 1.