AP EAMCET201823 Apr 2018Evening ShiftMathematicsComplex NumberActual
P is a point denoting z in the argand diagram and if z-i z-1 is always purely imaginary, then locus of P is
Options
- Athe circle with centre ( 1 2 , 1 2 ) and radius 1 2
- Bthe circle with centre (- 1 2 ,- 1 2 ) and radius 1 2
- Cthe points on the circle with centre ( 1 2 , 1 2 ) and radius 1 2 , excluding the points (1,0) and (0,1)
- Dthe points on the circle with centre (- 1 2 ,- 1 2 ) and radius 1 2 , excluding the origin
Correct answer
C. the points on the circle with centre ( 1 2 , 1 2 ) and radius 1 2 , excluding the points (1,0) and (0,1)
Step-by-step solution
Let z=x+i y , then z-i z-1 = x+i(y-1) (x-1)+i y So, Re ( z-i z+1 )= x(x-1)+y(y-1) (x-1)^2+y^2 z-i z+1 is purely imaginary, So, Re ( z-i z+1 )=0 x^2+y^2-x-y=0 , and it is a circle with centre ( 1 2 , 1 2 ) and radius 1 2 , excluding the points (1,0) and (0,1) .