If z=x+iy(x,y∈R,x≠−1/2) the number of values of z satisfying |z|n=z2|z|n−2+z|z|n−2+1⋅(n∈N,n>1) is

The given equation is|z|n=z2+z|z|n−2+1⇒z2+z is real ⇒z2+z=z¯2+z¯⇒(z−z¯)(z+z¯+1)=0⇒z=z¯=x as z+z¯+1≠0(x≠−1/2)Hence, the given equation reduces toxn=xn+x|x|n−2+1⇒x|x|n−2=−1⇒x=−1So, the number of solutions is 1.