Question 3449
 Let z = a+bi where a,b are real.
 If z/(z-i) = c for some real c.
 then a + b i = c(z-i)
 or a + b i = c(a+(b-1)i)
 or a + b i = ca + c(b-1)i
 So, a=ca and b = c(b-1)

 a=ca implies a(c-1) = 0 Hence, a=0 or c =1

 But, if c = 1, then b = c(b-1) implies b = b-1 impossible.

 Hence, we see that a = 0, z = bi 
 If b = 0, then z = 0 and c = 0
  and so z is imaginary or zero.

 Kenny