Question

If `z^2/(z-1)` is always real, what is the locus of `z`?

Answer 1

See below.

Explanation:

Making `z = x + i y` we have

`(x+ i y)^2/(x+i y -1) = 1 + x + (x-1)/((x-1)^2 + y^2)+i(y - y/((x-1)^2 + y^2))`

now we need `y - y/((x-1)^2 + y^2)=0 rArr y(1 - 1/((x-1)^2 + y^2))=0` and we have the conditions

`{(y = 0),((x-1)^2+y^2 = 1):}`

so the locus is a line `y = 0` and a circle `(x-1)^2+y^2= 1`