`|z-w|^2+|z-w^2|^2=lambda`......(1)
`barz xx z=|z|^2`
so (1) turns into =>`bar|z-w|xx |z-w| +bar|z-w^2| xx |z-w^2|=lamda`
`|barz-barw|xx |z-w| +|barz-bar(w^2)| xx |z-w^2|=lamda`
opening and simplifying we get=>`2zbarz-(w+w^2)barz-(barw+bar(w^2))z+w^2+(w^2)^2=lamda`
`2|z|^2-(w+w^2)barz-(barw+bar(w^2))z+|w^2|+(|w^2|)^2=lamda`
`|w|=1 and 1+w+w^2=0`
substituting these we get =>`2|z|^2+barz+z+2=lamda`
if `z=x+iotay+barz=x-iotay ` then z+barz=2x
substituting values=>`x^2+y^2+x+((2-lamda)/2)=0`
centre of circle=`(-1/2,0)`
radius=`sqrt(1/4-(2-lamda)/2)=sqrt((lamda)/2-3/4` for circle to exist `radius>=0`
`(lamda>=3/2)` so `lamda` belongs to`(3/2,oo)`
