If `|z-2-i|=|z|sin(pi/4-a r g z)|` , where `i=sqrt(-1)` ,then locus of z, is

Options

A. pair of straight lines
B. circle of radius `sqrt(2)`
C. parabola
D. ellipse

Answer

C

Explanation

Let `z=x+iy`. Then,
`|z-2+i|=|z||sin(pi/4-"arg"(z))|`
`rArr |(x-2)+i(y+1)|=sqrt(x^(2)+y^(2))|sin(tan^(-1)-tan^(-1)y/x)|`
`rArr |(x-2)+i(y+1)|=sqrt(x^(2)+y^(2))|sin(sin^(-1)(x-y)/(sqrt(2(x^(2)+y^(2))))|`
`rArr sqrt((x-2)^(2)+(y+1)^(2))=|(x-y)/sqrt(2)|`
Distance of `(x,y)` from `(2,-1)` is same as the distance of `(x,y)` from `x-y=0`.
Hence, the locus of `(x,y)` is a parabola.