If A=z=x+iy/real part of (z¯-1)/(z-i)=2 then the locus of the point Px,y in the Cartesian plane is

Question

MARKS Content Team · Accepted Answer

Correct answer: (A) a pair of lines passing through -1,1. Given that Real part of (z¯-1)/(z-i) = 2 here z = x+iy ⇒z¯ = x-iy ⇒(z¯-1)/(z-i) = (x-iy-1)/(x+iy-i)=( x-1-iy)/(x+iy-1) ⇒(z ¯-1)/(z-i) = (x-1-iy ×x-iy-1)/(x^2+y-1^2) ⇒(z ¯-1)/(z-i) = (x^2-x-y-y^2-ixy-x-y+1+xy)/(x^2+y-1^2) Real part=2 ⇒x^2-x+y-y^2=2 x^2+1+y^2-2y ⇒x^2+3y^2+x-5y+2=0 The above represents a pair of straight lines passing through -1,1.

If $A = \{z = x + i y / real part of \frac{\bar{z} - 1}{z - i} = 2\}$ then the locus of the point $P (x, y)$ in the Cartesian plane is

Solution

Relevant Questions

If A=z=x+iy/real part of z¯-1z-i=2 then the locus of the point Px,y in the Cartesian plane is

Solution

Relevant Questions

If $A = \{z = x + i y / real part of \frac{\bar{z} - 1}{z - i} = 2\}$ then the locus of the point $P (x, y)$ in the Cartesian plane is